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THE  MANUFACTURE 

OF 

PULP  AND  PAPER 


VOLUME  I 


Pulp  and  Paper  Manufacture 

IN  FIVE  VOLUMES 

An  Official  Work  Prepared 
under  the   direction  of  the 

Joint  Executive  Committee  of  the 

Vocational  Education  Committees  of  the 

Pulp  and  Paper  Industry  of  the 

United  States  and  Canada 


Vol.     I — Mathematics,     How    to    Read 
Drawings,  Physics. 
II — Mechanics     and     Hydraulics, 
Electricity,  Chemistry. 

Ill — Preparation  of  Pulp. 
IV,  V — Manufacture  of  Paper. 


THE 

MANUFACTURE 

OF 

PULP  AND  PAPER 

A   TEXTBOOK  OF  MODERN  PULP 
AND     PAPER     MILL     PRACTICE 

Prepared   Under    the    Direction    of    the   Joint   Executive 

Committee  on  Vocational  Education  Representing 

the   Pulp    and    Paper    Industry    of    the 

United  States   and   Canada 


VOLUME  I 

Arithmetic,    Elementary  Applied  Mathematics, 
How  to  Read  Drawings,     Elements  of  Physics 

BY 

J.  J.  CLARK,  M.E. 


Fikst  Edition 
Third  [mprebbion 


McGRAW-HILL  BOOK  COMPANY,  Inc. 
NEW  YORK:    370   SEVENTH    AVENUE 

LONDON:    6  &  8  BOUVER1E  ST.,  E.  C.  4 
1921 


Copyright,  1921,  by  the 

Joint  Executive  Committee  of  the  Vocational  Education  Committees 

of  the  Pulp  and  Paper  Industry. 


All  Rights  Reserved, 
Including  Those  of  Translation. 


PRINTED   IN"   THE    I'NITED   STATES    OF   AMERICA 


IAPLE    PRESS    -    YORK     PA 


PREFACE 


In  numerous  communities  where  night  schools  and  extension 
classes  have  been  started  or  planned,  or  where  men  wished  to 
study  privately,  there  has  been  difficulty  in  finding  suitable 
textbooks.  No  books  were  available  in  English,  which  brought 
together  the  fundamental  subjects  of  mathematics  and  element- 
ary science  and  the  principles  and  practice  of  pulp  and  paper 
manufacture.  Books  that  treated  of  the  processes  employed 
in  this  industry  were  too  technical,  too  general,  out  of  date,  or  so 
descriptive  of  European  machinery  and  practice  as  to  be  unsuit- 
able for  use  on  this  Continent.  Furthermore,  a  textbook  was 
required  that  would  supply  the  need  of  the  man  who  must  study 
at  home  because  he  could  not  or  would  not  attend  classes. 

Successful  men  are  constantly  studying;  and  it  is  only  by 
studying  that  they  continue  to  be  successful.  There  are  many 
men,  from  acid  maker  and  reel-boy  to  superintendent  and  mana- 
ger, who  want  to  learn  more  about  the  industry  that  gives  them  a 
livelihood  and  by  study  to  fit  themselves  for  promotion  and  in- 
creased earning  power.  Pulp  and  paper  makers  want  to  under- 
stand the  work  they  are  doing — the  how  and  why  of  all  the 
various  processes.  Most  operations  in  this  industry  are,  to  some 
degree,  technical,  being  essentially  either  mechanical  or  chemical. 
It  is  necessary,  therefore,  that  the  person  who  aspires  to  under- 
stand these  processes  should  obtain  a  knowledge  of  the  under- 
lying laws  of  Nature  through  the  study  of  the  elementary  sciences 
and  mathematics,  and  be  trained  to  reason  clearly  and  logically. 

After  considerable  study  of  the  situation  by  the  Committee 
on  Education  for  the  Technical  Section  of  the  Canadian  Pulp 
and  Paper  Association  and  the  Committee  on  Vocational  Educa- 
tion for  the  Technical  Association  of  the  (U.  S.)  Pulp  and  Paper 
Industry,  a  joint  meeting  of  these  committees  was  held  in  Buffalo 

v 


vi  PREFACE 

in  September,  1918,  and  a  Joint  Executive  Committee  was  ap- 
pointed to  proceed  with  plans  for  the  preparation  of  the  text,  its 
publication,  and  the  distribution  of  the  books.  The  scope  of  the 
work  was  denned  at  this  meeting,  when  it  was  decided  to  provide 
for  preliminary  instruction  in  fundamental  Mathematics  and 
Elementary  Science,  as  well  as  in  the  manufacturing  operations 
involved  in  modern  pulp  and  paper  mill  practice. 

The  Joint  Educational  Committee  then  chose  an  Editor, 
Associate  Editor,  and  Editorial  Advisor,  and  directed  the  Editor 
to  organize  a  staff  of  authors  consisting  of  the  best  available  men 
in  their  special  lines,  each  to  contribute  a  section  dealing  with  his 
specialty.  A  general  outline,  with  an  estimated  budget,  was 
presented  at  the  annual  meetings  in  January  and  February,  1919, 
of  the  Canadian  Pulp  and  Paper  Association,  the  Technical 
Association  of  the  Pulp  and  Paper  Industry  and  the  American 
Paper  and  Pulp  Association.  It  received  the  unanimous  approval 
and  heart}'  support  of  all,  and  the  budget  asked  was  raised  by 
an  appropriation  of  the  Canadian  Pulp  and  Paper  Association 
and  contributions  from  paper  and  pulp  manufacturers  and  allied 
industries  in  the  United  States,  through  the  efforts  of  the 
Technical  Association  of  the  Pulp  and  Paper  Industry. 

To  prepare  and  publish  such  a  work  is  a  large  undertaking; 
its  successful  accomplishment  is  unique,  as  evidenced  by  these 
volumes,  in  that  it  represents  the  cooperative  effort  of  the  Pulp 
and  Paper  Industry  of  a  whole  Continent. 

The  work  is  conveniently  divided  into  sections  and  bound  into 
volumes  for  reference  purposes;  it  is  also  available  in  pamphlet 
form  for  the  benefit  of  students  who  wish  to  master  one  part 
at  a  time,  and  for  convenience  in  the  class  room.  This  latter 
arrangement  makes  it  very  easy  to  select  special  courses  of 
study;  for  instance,  the  man  who  is  specially  interested,  sa}',  in 
the  manufacture  of  pulp  or  in  the  coloring  of  paper  or  in  any 
oilier  special  feature  of  the  industry,  can  select  and  stud}-  the 
special  pamphlets  bearing  on  those  subjects  and  need  not  study 
others  not  relating  particularly  to  the  subject  in  which  he  is 
interested,  unless  he  so  desires.  The  scope  of  the  work  enables 
the  man  with  but  little  education  to  study  in  the  most  efficient 
manner  the  preliminary  subjects  that  are  necessary  to  a 
thorough  understanding  of  the  principles  involved  in  the  manu- 
facturing processes  and  operations;  these  subjects  also  afford  an 
excellent  review  and  reference  textbook  to  others.     The  work 


PREFACE  vii 

is  thus  especially  adapted  to  the  class  room,  to  home  .study, 
and  for  use  as  a  reference  book. 

It  is  expected  that  universities  and  other  educational  agencies 
will  institute  correspondence  and  class  room  instruction  in 
Pulp  and  Paper  Technology  and  Practice  with  the  aid  of  these 
volumes.  The  aim  of  the  Committee  is  to  bring  an  adequate 
opportunity  for  education  in  his  vocation  within  the  reach  of 
every  one  in  the  industry.  To  have  a  vocational  education 
means  to  be  familiar  with  the  past  accomplishments  of  one's  trade 
and  to  be  able  to  pass  on  present  experience  for  the  benefit  of 
those  who  will  follow. 

To  obtain  the  best  results,  the  text  must  bo  diligently  studied; 
a  few  hours  of  earnest  application  each  week  will  be  well  repaid 
through  increased  earning  power  and  added  interest  in  the  daily 
work  of  the  mill.  To  understand  a  process  fully,  as  in  making 
acid  or  sizing  paper,  is  like  having  a  light  turned  on  when  one 
has  been  working  in  the  dark.  As  a  help  to  the  student,  many 
practical  examples  for  practice  and  study  and  review  questions 
have  been  incorporated  in  the  text ;  these  should  be  conscientiously 
answered. 

The  Editor  extends  his  sincere  thanks  to  the  Committee  and 
others,  who  have  been  a  constant  support  and  a  source  of  in- 
spiration and  encouragement;  he  desires  especially  to  mention 
Mr.  George  Carruthers,  Chairman,  and  Mr.  R.  8.  Kellogg, 
Secretary,  of  the  Joint  Executive  Committee;  Mr.  J.  J.  Clark, 
Associate  Editor,  Mr.  T.  J.  Foster,  Editorial  Advisor,  and  Mr. 
John  Erhardt  of  the  McGraw-Hill  Book  Company,   Inc. 

The  Committee  and  the  Editor  have  been  generously  assist ed 
on  every  hand;  busy  men  have  written  and  reviewed  manuscript, 
and  equipment  firms  have  contributed  drawings  of  great  value 
and  have  freely  given  helpful  service  and  advice.  Among  these 
kind  and  generous  friends  of  the  enterprise  are:  Mr.  O.  Bache- 
Wiig,  Mr.  James  Beveridge,  Mr.  J.  Brooks  Beveridge,  Mr.  H. 
P.  Carruth,  Mr.  Martin  L.  Griffin,  Mr.  H.  R.  Harrigan,  Mr. 
Arthur  Burgess  Larcher,  Mr.  J.  O.  Mason,  Mr.  Elis  Olsson,  Mr. 
George  K.  Spence,  Mr.  Edwin  Sutermeister,  Mr.  F.  G.  Wheeler, 
and  American  Writing  Paper  Co.,  Dominion  Engineering  Works, 
E.  I.  Dupont  de  Nemours  Co.,  F.  C.  Huyck  &  Sons,  Hydraulic 
Machinery  Co.,  Improved  Paper  Machinery  Co.,  E.  D.  Jones 
&  Sons  Co.,  A.  D.  Little,  Inc.,  National  Aniline  and  Chemical 
Works,  Process  Engineers,   Pusey  &  Jones  Co.,  Rice,  Barton  & 


viii  PREFACE 

Falcs  Machine  and  Iron  Works,  Ticonderoga  Paper  Co..  Waterous 
Engine  Works  Co.,  and  many  others. 

J.  Newell  Stephenson, 

Editor 
For  the 
Joint  Executive  Committee  oh  Vocational  Education, 

George  Carruthers.  Chairman,  R.   B.  Kellogg,  Secretary, 

T.  L.  Crossley,     G.  E.  Williamson,         C.  P.  Win-slow. 

Representing  the  Technical  Sec-  Representing  the  Technical  As- 
tion  of  the  Canadian  Pulp  and  Paper  sociation  of  the  (U.  S.)  Pulp  and 
Association  Paper  Industry. 

T.  L.  Crossley,  Chairman,  George  E.  Williamson,  Chairman, 

George  Carruthers,  Hugh  P.  Baker, 

A.  P.  Costigane.  Henry  J.  Guild, 

Dan  Dayerin,  R.  S.  Kellogg, 

C.  Nelson  Gain,  Otto  Kress, 

J.  N.  Stephenson.  W.  S.  Lucey, 

C.  P.  Winslow. 


CONTENTS 


Page 

Preface v 

SECTION  1 
Arithmetic 

PART  I 

Notation  and  Enumeration 1-10 

Addition  and  Subtraction 11-21 

Multiplication  and  Division 22-34 

Some  Properties  of  Numbers 34-39 

Examination  Questions 41-42 

PART  II 

Greatest  Common  Divisor  and  Least  Common  Multiple  .    .    .  43-47 

Fractions 48-64 

Involution 64-67 

Decimals  and  Decimal  Fractions 67-76 

Signs  of  Aggregation 76-77 

Ratio  and  Proportion 77-85 

Examination  Questions 87-88 

PART  III 

Square  Root '.    .    .    .  89-98 

Percentage 98-106 

Compound  Numbers 106-113 

The  Metric  System 113-124 

The  Arithmetical  Mean 125-126 

Examination  Questions 127-128 

SECTION  2 
Elementary  Applied  Mathematics 

PART  I 

Mathematical  Formulas 1-4 

Algebraic     Addition,     Subtraction,     Multiplication,     and 

DrvisiON 4-18 

Equations 18-27 

Accuracy  in  Calculation 27-33 

Approximate  Method  for  Finding  Roots 33-39 

Examination  Questions 41-42 

ix 


x  CONTENTS 

PART  II 

Paoe 

Mensuration  of  Plane  Figures 43-47 

Triangles 50-60 

Quadrilaterals 61-66 

Regular  Polygons 66-68 

The  Circle 68-89 

Inscribed  and  Circumscribed  Polygons 89  96 

Examination  Questions 97-100 

PART  III 

Mensuration  of  Solids 101-130 

Similar  Figures 131-137 

Symmetrical  Figures 137-143 

Examination  Questions 145-146 


SECTION  3 
How  to  Read  Drawings 

Representing  Solids  on  Planes 1-11 

Special  Features  Pertaining  to  Drawings 12-23 

Reading  Drawings — Visualizing  the  Object 23-28 

Some  Examples  in  Reading  Drawing 28 


SECTION  4 
Elements  of  Physics 

PART  1 

Matter  and  Its.  Properties 1-6 

Motion  and  Velocity 6-9 

Force,  Mass  and  Weight 9-14 

Work  and  Energy 14-17 

Hydrostatics — Pascal's  Law 18-30 

Buoyancy  and  Specific  Gravity 31-39 

Capillarity 39-41 

Pneumatics 42-o4 

Examination  Questions 55-56 

PART  II 

Properties  of  Perfect  Gases 57-66 

Nat  rid:  of  .Measurement  of  Heat 66-80 

Change  of  State 81-90 

Light 90-117 

Examination  Questions 119-120 

Index 121-132 


SECTION  1 

ARITHMETIC 

PART  1 


NOTATION  AND  ENUMERATION 


DEFINITIONS 

1.  A  unit  is  the  standard  by  which  anything  is  measured.  For 
example  if  it  were  desired  to  measure  the  length  of  a  room,  it 
could  be  done  with  a  tape  line,  a  rule,  a  stick  of  some  convenient 
length,  or  anything  else  that  would  form  a  basis  of  comparison. 
Suppose,  for  example,  a  broom- 
stick were  used;  the  broomstick 
would  be  laid  on  the  floor,  with 
one  end  a  against  the  wall  at  one 
side  of  the  room  and  with  the 
other  end  b  extending  towards 
the  opposite  side  of  the  room, 
as  shown  in  Fig.  1.  The  point  b 
would  be  marked  and  the  broom- 
stick would  be  shifted  so  as  to 
occupy  the  position  be,  then  to 
the  position  cd,  and  so  on  until 
the  other  wall  was  reached. 
Each  of  the  lengths  ab,  be,  etc. 
is  equal  in  length  to  the  broomstick,  which  is  in  this  case  the  unit 
of  length.  The  word  that  denotes  how  many  times  the  room  is 
longer  than  the  broomstick  is  member.  In  other  words  number 
means  an  aggregation,  or  collection,  of  units;  a  number  may 
also  mean  a  single  unit  or  a  part  of  a  unit — it  refers  in  all  cases  to 
the  complete  measurement. 

As  another  example,  consider  the  compensation  that  a  man 
receives  for  his  work.  For  working  a  certain  number  of  hours  he 
receives  a  certain  number  of  dollars;  here  two  units  are  involved, 


a 

ft 

c 

a 

e 

/ 

i 

; 

■>       t 

: 

Pro.  1. 


2  ARITHMETIC  §1 

hours  and  dollars.  The  unit  of  hours  is  established  or  arrived  at 
by  dividing  or  splitting  up  a  day  (another  unit)  into  twenty-four 
equal  parts  and  calling  one  of  these  parts  one  hour.  The  unit  of 
money  is  established  by  law;  it  is  called  in  the  United  States  of 
America  one  dollar,  and  is  equivalent  in  value  to  about  twenty- 
three  and  one-fourth  grains  (another  unit)  of  pure  gold.  Canada 
has  a  unit  of  the  same  name  and  of  practically  the  same  value. 

2.  Three  different  kinds  of  units,  according  to  their  origin, 
have  been  mentioned,  the  broomstick,  which  may  be  called  an 
expediency  unit;  the  hour,  which  is  an  international  unit  for 
measuring  time;  and  the  dollar,  wrhich  is  a  legal  unit,  or  one  estab- 
lished by  law.  According  to  their  use,  units  have  many  names, 
and  anything  that  can  be  expressed  as  a  certain  number  of  units 
is  called  a  quantity.  The  science  that  treats  of  quantity  and  its 
measurement  is  called  mathematics. 

3.  The  difference  between  quantity  and  number  is  this:  quantity 
is  a  general  term  and  is  applied  to  anything  that  can  be  measured 
or  expressed  in  units;  number  is  simply  a  term  applied  to  a  unit 
or  a  collection  of  units  and  is  usually  restricted  to  expressions 
containing  only  figures. 

4.  The  act  or  process  by  which  numbers  are  used  to  reckon, 
count,  estimate,  etc.  is  called  computation  or  calculation. 

5.  Arithmetic  is  that  branch  of  mathematics  which  treats  of 
numbers  and  their  use  in  computation. 

Numbers  are  divided  into  two  general  classes,  according  to  their 
signification:  abstract  numbers  and  concrete  numbers. 

6.  An  abstract  number  is  one  not  applied  t8  any  object  or 
quantity,  as  three,  twelve,  one  hundred  seventeen,  etc. 

7.  A  concrete  number  is  one  relating  to  a  particular  kind  of 
object  or  quantity,  as  five  dollars,  eleven  pounds,  seven  hours,  etc. 

According  to  their  units,  numbers  are  also  divided  into  two 
general  classes: 

8.  Like  numbers  are  numbers  having  the  same  units:  for  ex- 
ample, seven  hours,  twelve  hours,  and  three  hours  are  like  numbers, 
since  they  all  have  the  same  unit — one  hour;  fourteen,  seventeen 
and  twenty  are  also  like  numbers,  the  unit  of  each  being  one.  All 
abstract  numbers  are  like  numbers. 

9.  Unlike  numbers  are  numbers  having  different  units;  for 
example,  five  dollars,  eight  hours,  and  ten  pounds  are  unlike 


§1  NOTATION  AND  ENUMERATION  3 

numbers  since  their  units  are  unlike,  being  respectively  one  dollar, 
one  hour,  and  one  pound. 


NOTATION 

In  order  that  numbers  may  be  recorded,  some  method  of  writing 
and  reading  them  must  be  devised. 

10.  Notation  is  the  word  used  to  express  the  act  of  and  the 
result  obtained  by  writing  numbers. 

11.  Numeration  is  the  act  or  process  of  reading  numbers  that 
have  been  written. 

12.  Notation  is  accomplished  in  three  ways:  (1)  by  words;  (2) 
by  letters;  (3)  by  figures.  The  first  method,  by  words,  is  never 
used  in  computation,  only  the  second  and  third  methods  having 
ever  been  employed  for  this  purpose.  The  second  method,  called 
the  Roman  notation,  is  very  seldom  used  at  the  present  time,  its 
only  use  is  for  numbering,  as  the  indexes,  chapters,  etc.  of  books, 
and  in  a  few  cases  for  dates.  The  third  method,  the  Arabic 
notation,  employs  certain  characters,  called  figures,  and  is  in 
universal  use  today. 


THE  FIRST  OR  WORD  METHOD 

13.  This  is  really  only  the  names  of  the  different  numbers. 
The  first  twelve  numbers  have  distinct  names;  beyond  these  all 
are  formed  according  to  a  definite  system  that  is  readily  learned. 
The  names  of  the  first  twelve  numbers  are:  one,  two,  three, 
four,  five,  six,  seven,  eight,  nine,  ten,  eleven,  twelve.  The  next 
seven  numbers  all  end  in  teen,  which  means  and  ten,  the  first  part  of 
the  word  being  derived  from  the  words  three,  four,  etc.  up  to  nine; 
they  are  thirteen,  fourteen,  fifteen,  sixteen,  seventeen,  eighteen, 
and  nineteen.  Thirteen  means  three  and  ten,  fourteen  means 
four  and  ten,  etc.  The  name  of  the  next  number  is  twenty,  and 
the  names  of  the  next  nine  numbers  are  all  compound  words, 
having  for  their  first  element  the  word  twenty  and  for  their  second 
element  the  words  one,  two,  etc.  to  nine.  Thus,  twenty-one, 
twenty-two,  etc.  to  twenty-nine.  The  word  twenty  means  two 
tens.  The  next  number  is  thirty,  which  means  three  tens,  and  the 
next  nine  are  thirty-one,  thirty-two,  etc.  to  thirty-nine.  Then 
follow  forty,  fifty,  sixty,  seventy,  eighty,  and  ninety,  with  their 


4  ARITHMETIC  §1 

combinations  of  one,  two,  etc.  to  nine.  The  number  following 
ninety-nine  is  one  hundred.  The  numbers  are  then  repeated, 
with  the  element  otic  hundred  added,  one-hundred-one,  one- 
hundred-two,  etc.,  one-hundred-twenty-one,  one-hundred-twenty- 
two,  etc.  to  one-hundred-ninety-nine,  the  next  number  being 
two-hundred.  The  notation  is  continued  in  this  manner  to  nine- 
hundred-nincty-nine,  the  next  number  being  one-thousand;  then 
follow  one-thousand-one,  one-thousand-two,  etc.,  one-thousand- 
one-hundred-one,  one-thousand-one-hundred-two,  etc.  to  nine- 
thousand-nine-hundred-ninety-nine,  the  next  number  being  ten- 
thousand.  Then  follow  ten-thousand-one,  etc.  to  nine-hundred- 
thousand-nine-hundred-ninety-nine,  the  next  number  being  called 
one-million.  The  notation  is  then  carried  on  one-million-one,  etc. 
up  to  one-thousand-million,  which  in  the  United  States,  France, 
and  the  majority  of  other  countries  is  called  a  billion;  a  thousand 
billion  is  a  trillion,  etc.  In  Great  Britain  and  a  few  other  coun- 
tries, a  billion  is  a  million-million,  a  trillion  a  million-billion,  etc. 


THE  ROMAN  NOTATION 

14.  The  Roman  notation  uses  the  letters  I,  V,  X,  L,  C,  D,  and 
M.  The  values  of  the  letters  when  standing  alone  are:  I  is  one, 
V  is  five,  X  is  ten,  L  is  fifty,  C  is  one-hundred,  D  is  five-hundred, 
M  is  one-thousand.  Formerly  other  letters  were  used,  but  these 
are  all  that  are  employed  now.  Numbers  are  expressed  by  letters 
in  accordance  with  the  following  principles: 

I.  Repeating  a  letter  repeats  its  value. 

Thus,  I  is  one,  II  is  two,  III  is  three,  XX  is  two  tens  or  twenty, 
XXX  is  three  tens  or  thirty,  CC  is  two  hundred,  MM  is  two- 
thousand,  etc. 

V,  D,  and  L  are  never  repeated;  only  I,  X,  C,  and  M  are  ever 
used  more  than  once  in  any  number,  except  when  they  precede  a 
letter  of  higher  value. 

II.  7/  a  letter  precedes  one  of  greater  value,  their  difference  is 
denoted;  if  it  follows  a  letter  of  greater  value,  their  sum  is  denoted. 

Thus,  IV  is  four,  VI  is  six,  IX  is  nine,  XII  is  twelve,  XL  is  forty, 
LX  is  sixty.  XC  is  ninety,  etc.  Only  one  letter  of  lower  value  is 
allowed  to  precede  one  of  higher  value,  and  in  general,  the  only 
letters  used  are  I  and  X.  Further,  I  precedes  only  V  and  X,  and 
X  precedes  only  L  and  C.     Occasionally,  however,  C  is  used  in 


§1  NOTATION  AND  ENUMERATION  5 

this  manner  before  M ;  as  for  example,  in  the  date  nineteen-hun- 
dred-four,  when  instead  of  writing  MDCCCCIV  the  shorter  form 
MCMIV  is  used.  Pope  Leo  XIII  wrote  the  date  eighteen- 
hundred  ninety-five  MDCCCVC  instead  of  MDCCCXCV.  On 
the  Columbus  memorial  in  front  of  the  Union  Station,  Washington, 
D.  C,  the  year  of  his  birth  is  expressed  by  MCDXXXVI,  which, 
of  course,  is  read  fourteen-hundred-thirty-six. 

III.  A  bar  placed  over  a  letter  multiples  its  value  by  one-thousand. 

Thus,  X  is  ten-thousand,  L  is  fifty-thousand,  XCDXVII  is 
ninety-thousand  five-hundred  seventeen.  In  the  last  expression, 
it  might  be  thought  that  D  was  preceded  by  X  and  C;  this  is  not 
the  case,  however,  as  XC  is  treated  as  a  single  letter  having  a 
value  of  ninety-thousand. 

The  following  table  illustrates  the  foregoing  principles  and 
applications  of  the  Roman  notation : 

VIII  is  eight 

XII  is  twelve 

XVIII  is  eighteen 

XXIX  is  twenty-nine 

XXXV  is  thirty-five 

XLIV  is  forty-four 

LXXVI  is  seventy-six 

XCII  is  ninety-two 

CXLIV  is  one-hundred  forty-four 

CCCCL  is  four-hundred  fifty 

IXLX  is  nine-thousand  sixty 

C  is  one-hundred-thousand 

D  is  five-hundred-thousand 

M  is  one-million 

THE  ARABIC  NOTATION 
15.  The    Arabic    notation    employs    ten    characters,    called 
figures,  to  represent  numbers;  these  characters  together  with 
their  names  are : 

0  12345  6  789 

naught  one        two     three    four      five       six      setien     eight     nine 

cipher 
zero 

The  first  figure  (0)  is  called  naught,  cipher,  or  zero,  and  has  no 
value.  The  other  nine  figures  arc  called  digits,  and  each  has 
the  value  denoted  by  its  name. 


6  ARITHMETIC  §1 

16.  For  numbers  higher  than  nine  it  is  necessary  to  use  two  or 
more  figures  to  express  them.  Thus  ten  is  expressed  by  the  com- 
bination 10,  the  cipher  indicating  that  there  are  no  units  in  the 
right-hand  figure;  11  is  eleven,  12  is  twelve,  13  is  thirteen,  etc.  up 
to  19  or  nineteen.  All  these  expressions  mean  ten  and  one,  ten 
and  two,  etc.  Twenty  is  written  20,  twenty-one,  21,  etc.;  thirty 
is  written  30,  thirty-three,  33,  etc.;  and  so  on  up  to  99  or  ninety- 
nine.  One-hundred  is  written  100,  the  right-hand  cipher  mean- 
ing that  there  are  no  units  and  the  next  that  there  are  no  tens. 
Five-hundred-forty-two  is  written  542,  five-hundred-two  is  writ- 
ten 502,  five-hundred-forty  is  written  540.  In  a  similar  manner 
one-thousand  is  written  1000,  ten-thousand  is  written  10000, 
one-hundred-thousand  is  written  100000,  one-million  is  written 
1000000,  etc. 

17.  As  stated  before  the  cipher  has  no  value,  but  it  plays  a  very 
important  part  in  expressing  numbers  by  figures ;  it  not  only  indi- 
cates the  absence  of  a  digit  from  the  place  it  occupies,  but  it  also 
locates  the  digits  with  respect  to  one  another. 

Any  number,  as  for  example,  5642,  is  equivalent  to  5000  and 
600  and  40  and  2 ;  that  is,  in  this  case,  it  is  equivalent  to  four  sepa- 
rate numbers,  three  of  which  are  single  digits,  all  followed  by  one 
or  more  ciphers.  Such  a  number  might  be  written  5000  +  600 
+  40  +  2,  but  this  is  not  necessary,  since  the  ciphers  can  be 
omitted  without  impairing  the  result,  with  the  additional  ad- 
vantage of  saving  space  and  figures. 

18.  The  number  denoted  by  a  figure  in  any  whole  number  is 
determined  by  writing  the  figure  and  then  writing  after  it  as 
many  ciphers  as  there  are  figures  to  the  right  of  the  one  selected. 
Thus,  in  the  number  2645893,  the  number  denoted  by  the  figure 
6  is  600000  or  six-hundred-thousand ;  the  number  denoted  by  the 
figure  5  is  5000  or  five-thousand;  etc. 


NUMERATION 

19.  For  convenience  numbers  are  divided  into  orders  and 
periods.  A  figure  belongs  to  the  first  order  when  it  occupies  the 
units  place;  it  belongs  to  the  second  order  when  it  occupies  the 
tens  place;  etc.  In  the  number  2645893,  2  belongs  to  the  seventh 
order,  6  to  the  sixth  order,  etc.  Numbers  are  divided  by  commas 
into  periods  of  three  figures  each,  beginning  with  the  first  order, 


§1 


NOTATION  AND  ENUMERATION 


to  assist  in  reading  them;  the  periods  are  named  according  to  the 
name  of  the  lowest  order  in  the  periods.  Thus,  the  number 
506,273,985,114  contains  four  full  periods;  beginning  at  the  right 
and  going  to  the  left,  the  names  of  the  periods  are,  respectively, 
units,  thousands,  millions,  billions.  The  names  of  periods  and 
orders  represented  by  the  figures  in  the  above  number  are 
conveniently  shown  in  the  table  below. 


Fourth  Period. 
Billions. 

Third  Period. 
Millions. 

Second  Period. 
Thousands. 

Ftrsi  Period. 
Units. 

CO 

C 

o 

m 

73 
« 

E 

73 
3 

w 

CO 

73 

E 

o 

.3 

1 

CO 

H 

CO 

s 

« 
■ 

s 

9 

H 

CD 
73 
E 

O 

,3 

a 
$ 

> 

CO 

3 
O 

« 

c 

CO 

73 
E 

O 

■3 

+3 

c 
a> 
H 

CO 

a 
_o 

3 

73 

0) 
Fh 
73 

C 
3 

w 

(J 

CO 

n3 
E 

O 

.3 
3 

CO 

a 
o 

3 
■ 

a 

CO 

H 

tJ 
a) 

o 

-<■» 

,4 

cd 

a 

o 

3 

M 

CO 

13 

Ei 

o 

CO 

> 

CO 

co 

73 

C 
(S 
CO 

O 

P 
■ 

73 
O 

(- 
73 

a 

M 

CO 
73 
J- 

o 

X 

o3 

CO 
73 

3 

3 

CO 

O 

P 
i 

3 

CO 

H 

CO 
73 

E 

o 

.3 
-ta 

CO 
73 

3 

33 
CQ 

6 

-3 
P 

u 
CD 

73 

E 

o 

-1-3 

u 

3 

co 
73 
CO 
(h 
73 
3 
3 

w 

CO 
73 

fi 

o 

73 

E 

P 

CO 

3 
Oi 

H 

c 

CO 

73 

E 

o 

73 

3 
O 
V 

CO 

CO 

45 

"3 

M 

CO 
73 
M 

o 

-1-3 
50 

ft 

0 


6, 


3, 


5, 


In  pointing  off  these  figures,  begin  at  the  right-hand  figure  and 
count — units,  tens,  hundreds;  the  next  group  of  three  figures  is 
thousands;  therefore,  insert  a  comma  (,)  before  beginning  with 
them.  Beginning  at  the  figure  5,  say  thousands,  ten  thousands, 
hundred  thousands,  and  insert  another  comma;  next  read  millions, 
ten  millions,  hundred  millions  (insert  another  comma) ;  lastly,  read 
billions,  ten  billions,  hundred  billions. 

20.  A  number  is  read  by  beginning  at  the  left  and  reading  the 
figures  composing  each  period  as  though  it  formed  a  number  by 
itself  and  affixing  the  name  of  the  period,  taking  each  period  in 
succession  from  left  to  right.  Thus,  the  above  number  is  read: 
five-hundred-six  billion  two-hundred  seventy-three  million  nine- 
hundred-eighty-five  thousand  one-hundred-fourteen.     Note  that  the 


8  ARITHMETIC  §1 

name  of  the  units  period  is  not  pronounced — it  is  always 
understood. 

Reading  a  line  of  figures  in  this  manner  is  called  numeration; 
and  when  the  numeration  is  changed  back  to  figures,  it  is  called 
notation. 

For  instance,  the  writing  of  the  following  figures, 

72,584,623 

would  be  the  notation,  and  the  numeration  would  be  seventy-two 
m  ill  ion  fivc-h  undred-eighty-four  thousand  six-hundred-twenty-ihree. 

21.  Note. — It  is  customary  to  leave  the  "s"  off  the  words  millions,  thou- 
sands, etc.  in  cases  like  the  above,  both  in  speaking  and  writing. 

22.  Consider  the  number  500,000.  If  the  5,  which  is  here  a 
figure  of  the  sixth  order,  be  moved  one  place  to  the  right,  so  as  to 
occupy  the  fifth  order,  the  number  becomes  50,000,  which  is  only 
one-tenth  as  large  as  before;  while  if  the  5  be  moved  one  place  to 
the  left,  so  as  to  occupy  the  seventh  order,  the  number  is  ten 
times  larger.  In  other  words,  moving  a  figure  to  the  right  de- 
creases the  number  it  represents  ten  times  for  each  order  it 
passes  into;  and  moving  the  figure  to  the  left  increases  the  number 
represented  by  the  digit  ten  times  for  each  order  passed  into. 
Thus,  5  is  one-tenth  of  50  and  50  is  ten  times  5;  50  is  one-tenth  of 
500  and  500  is  ten  times  50;  etc. 

23.  Integers  and  Decimals.  Any  figure,  say  5,  represents  the 
number  of  units  signified  by  its  name,  when  standing  alone. 
Moving  it  one  place  to  the  left,  it  represents  five  tens  or  fifty,  and 
is  written  50;  moving  it  one  place  to  the  right,  it  represents  five- 
tenths,  and  is  written  0.5.  Moving  the  figure  two  places  to  the 
left  of  its  first  position,  it  represents  five  hundred,  written  500;  if 
moved  two  places  to  the  right,  it  represents  five  hundredths, 
written  0.05.  Now  0.5  means  that  this  number  is  only  one-tenth 
as  large  as  5,  and  0.05  means  that  this  number  is  only  one-tenth 
as  large  as  0.5  or  one-hundredth  as  large  as  5.  This  notation 
may  be  continued  to  the  right  as  far  as  desired. 

24.  The  dot  is  called  the  decimal  point;  it  is  used  to  point  out 
the  unit  figure,  and  the  part  of  the  number  to  the  right  of  the 
unit  figure  is  called  a  decimal.  All  numbers  not  having  any 
digits  bo  the  righl  of  their  unit  figures  (and  which  do  not  contain 
a  common  fraction)  are  called  whole  numbers  or  integers.  The 
numbers  dealt  with  heretofore  have  all  been  integers.     The  part 


§1  NOTATION  AND  ENUMERATION  9 

of  a  number  to  the  left  of  the  decimal  point  is  called  the  integral 
part  of  the  number. 

25.  Decimals  are  read  in  much  the  same  manner  as  integers. 
The  only  difference  is  that  after  the  number  has  been  read  as 
though  it  were  an  integer,  the  name  of  the  lowest  (right-hand) 
order  is  pronounced  with  the  letters  ths  added  to  the  name.  Thus, 
0.52976  is  read  fifty-two  thousand  nine-hundred-seventy-six 
hundred-thousandths;  0.529  is  read  five-hundred-twenty-nine 
thousandths. 

26.  In  using  decimals  be  careful  to  mark  the  units  place  by 
placing  the  decimal  point  immediately  to  the  right  of  the  figure  in  the 
units  place.  Then  continue  the  notation  to  the  right  of  the  units 
place  precisely  as  to  the  left  of  it.  The  resemblance  between  the 
names  of  the  places  on  the  two  sides  of  the  units  place  is  shown  by 
the  following  table,  in  which  the  differences  are  marked  by 
italics. 


T3 

a 

cj 

00 

3 
O 
X 
-to 

C 

<n 

3 

£3 

0J 

O 

s 

O 

«-. 

X 

l~ 

-a 

+j 

3 
O 
X 

T3 

a 

3 

cl 

4) 

e 

3 

XI 

+3 

+3 

X! 

"O 

c 

oo 

08 
CO 

■<e 

T3 

c 

c5 

3 
O 

X 

T3 
C 
03 
a> 

3 
o 
X 

3 
O 
X 

3 
O 

1 

S 

(- 

a 

3 
X 

-tJ 

27.  In  this  (the  Arabic)  system  of  notation,  each  place  or  order 
is  said  to  be  higher  than  any  place  to  the  right  of  it,  and  lower 
than  any  place  to  the  left  of  it.  Thus,  in  the  number  43.127,  the 
highest  place  is  tens,  and  the  lowest  place  is  thousandths. 

28.  In  writing  and  printing,  decimals  are  frequently  written 
.5,  .529  etc.  instead  of  0.5,  0.529,  etc.  The  latter  way  is  the 
better  and  more  correct,  and  the  student  is  advised  to  write  the 
cipher  in  all  cases;  for,  if  the  decimal  point  should  fail  to  print,  be 
indistinct,  or  through  carelessness  not  written,  the  cipher  will 
frequently  serve  to  distinguish  between  integers  and  decimals. 
It  will  be  found,  however,  that  the  cipher  has  frequently  been 
omitted  throughout  this  textbook  to  accustom  the  reader  to 
both  methods  of  writing  decimals. 

29.  A  number,  part  of  which  is  an  integer  and  part  a  decimal 
is  called  a  mixed  number.     A  mixed  number  is  read  by  reading 


10  ARITHMETIC  §1 

the  integral  part  first,  pronouncing  the  word  and,  and  then 
reading  the  decimal  part.  Thus,  the  number  given  in  Art.  26, 
or  543,210.12345  is  read  five-hundred-forty-three  thousand 
two-hundred-ten  and  twelve-thousand-three-hundred-forty-five 
hundred-thousandths. 

The  abbreviation  "Art."  means  Article,  and  refers  to  the 
numbered  paragraphs  or  sections  throughout  this  textbook.  The 
plural  is  written  Arts.;  thus,  Arts.  1-10,  means  articles  1  to  10, 
both  inclusive. 

Decimals  are  seldom  pointed  off  into  periods,  especially  in 
English-speaking  countries.  It  ma}'  be  done,  however,  (if 
desired)  in  the  same  manner  as  integers,  beginning  at  the  unit 
figure.  Thus,  0.0000000217  would  be  pointed  off  0.00,000,002,17. 
In  reading  this  decimal,  name  the  periods  until  the  last  is  reached 
and  then  name  the  order  of  the  right-hand  figure; — in  this  case, 
units,  thousandths,  millionths,  billionths,  ten-billionths; — this 
determines  the  name  of  the  lowest  order.  The  number  is  then 
read  at  once  as  two-hundred-seventeen  ten-billionths. 


EXAMPLES  FOR  PRACTICE 

Read  the  following  numbers  and  write  their  names: 

(1)  20103 

(2)  1964727 

(3)  7926.4867 

(4)  0.0078314 

Write  the  following  numbers: 

(5)  Eight-hundred-sixty-six  million    forty-six-thousand     seven-hundred- 
thirty-three. 

(6)  Six-hundred-ninety -one  and  four-thousand-five-hundred-twenty-three 
ten-thousandths. 

(7)  Seventeen    and    eighty-nine-thousand-two-hundrcd-scventy-six  mill- 
ionths. 


30.  The  Four  Fundamental  Processes. — The  processes  of 
addition,  subtraction,  multiplication  and  division  are  called 
fundamental  because  all  other  processes  employed  in  arithmetical 
computation  are  based  on  them;  in  other  words,  no  computation 
can  be  made  without  employing  one  or  more  of  these  processes. 


§1  ADDITION  AND  SUBTRACTION  11 

ADDITION  AND  SUBTRACTION 


ADDITION 

31.  Suppose  it  is  desired  to  ascertain  the  length  of  a  room,  and 
that  the  distance  from  one  end  to  the  edge  of  a  door  nearest  that 
end,  the  width  of  the  door,  and  the  distance  from  the  other  edge  of 
the  door  to  the  other  end  of  the  room  are  known  to  be  respect- 
ively, 6  feet,  3  feet,  and  8  feet;  then  without  getting  a  rule  or 
other  measure,  the  length  of  the  room  can  be  found  by  counting, 
starting  with  six,  counting  three  more,  getting  nine,  and  then 
counting  eight  more  after  nine,  getting  seventeen.  In  other 
words,  the  combined  lengths  of  6  feet,  3  feet,  and  8  feet  is  17  feet, 
and  the  numbers  6,  3,  and  8  taken  together  always  give  17.  The 
process  of  finding  a  single  number  equivalent  in  amount  to  two 
or  more  numbers  taken  together  is  called  addition,  and  the  single 
number  found  as  the  result  of  the  process  is  called  the  sum. 
Thus,  the  sum  of  6,  3,  and  8  is  17.  To  add  two  or  more  numbers 
is  to  find  their  sum. 

32.  The  method  just  described  for  finding  the  sum  of  several 
numbers  is  impracticable  except  for  very  small  numbers.  For 
example,  suppose  it  were  desired  to  find  the  sum  of  $2,904,  $976, 
and  $3,520;  not  only  would  this  take  a  very  long  time,  but  there 
would  also  be  great  danger  of  making  a  mistake.  A  method  for 
finding  the  sum  of  a  series  of  numbers  will  now  be  described. 

When  it  is  desired  to  indicate  that  several  numbers  are  to  be 
added,  they  are  written  in  a  row  and  separated  by  Greek  crosses. 
Thus,  the  fact  that  the  numbers  6,  3,  and  8  are  to  be  added  is 
indicated  as  follows:  6  +  3  +  8.  The  cross  +  is  called  the 
sign  of  addition  or  plus  or  plus  sign,  and  the  expression  would  be 
read,  six  plus  three  plus  eight.  If  it  is  also  desired  to  write  the 
sum  as  indicating  the  result  of  the  addition,  this  is  done  in  the 
following  manner. 

6  +  3  +  8  =  17 
The  sign  =  is  called  the  equality  sign,  it  is  read  equals  or  is 
equal  to,  and  signifies  that  the  result  of  the  operations  indicated 
on  one  side  is  exactly  equal  to  the  number  (or  to  the  result  of  a 
series  of  operations  indicated)  on  the  other  side.  For  example, 
5  +  2  +  10  =  17;  hence,  6  +  3  +  8  =  5  +  2+ 10  =  17.  This 
means  that  the  result  of  the  operation  6  +  3  +  8  is  equal  to  the 
result  of  the  operation  5  +  2  +  10  and  that  either  sum   is  17. 


12 


ARITHMETIC  §1 

17  is  read,  six  plus  three  plus 


An  expression  like  6  +  3  +  8 
eight  equals  seventeen. 

33.  Before  one  can  add  several  numbers  it  is  necessary  that 
the  sum  of  any  two  numbers  from  1  to  9  be  recognized  as  soon  as 
seen.     The  following  table  gives  the  sum  of  any  two  numbers 
ADDITION  TABLE 


1  and  1  is  2 

2  and  1  is  3 

3  and  1  is  4 

4  and  1  is  5 

1  and  2  is  3 

2  and  2  is  4 

3  and  2  is  5 

4  and  2  is  6 

1  and  3  is  4 

2  and  3  is  5 

3  and  3  is  6 

4  and  3  is  7 

1  and  4  is  5 

2  and  4  is  6 

3  and  4  is  7 

4  and  4  is  8 

1  and  5  is  6 

2  and  5  is  7 

3  and  5  is  8 

4  and  5  is  9 

1  and  6  is  7 

2  and  6  is  8 

3  and  6  is  9 

4  and  6  is  10 

1  and  7  is  8 

2  and  7  is  9 

3  and  7  is  10 

4  and  7  is  11 

1  and  8  is  9 

2  and  8  is  10 

3  and  8  is  11 

4  and  8  is  12 

1  and  9  is  10 

2  and  9  is  11 

3  and  9  is  12 

4  and  9  is  13 

1  and  10  is  11 

2  and  10  is  12 

3  and  10  is  13 

4  and  10  is  14 

1  and  11  is  12 

2  and  11  is  13 

3  and  11  is  14 

4  and  11  is  15 

1  and  12  is  13 

2  and  12  is  14 

3  and  12  is  15 

4  and  12  is  16 

5  and  1  is  6 

6  and  1  is  7 

7  and  1  is  8 

8  and  1  is  9 

5  and  2  is  7 

6  and  2  is  8 

7  and  2  is  9 

8  and  2  is  10 

5  and  3  is  8 

6  and  3  is  9 

7  and  3  is  10 

8  and  3  is  J 1 

5  and  4  is  9 

6  and  4  is  10 

7  and  4  is  11 

8  and  4  is  12 

5  and  5  is  10 

6  and  5  is  11 

7  and  5  is  12 

8  and  5  is  13 

5  and  G  is  11 

6  and  6  is  12 

7  and  6  is  13 

8  and  6  is  14 

5  and  7  is  12 

6  and  7  is  13 

7  and  7  is  14 

8  and  7  is  15 

5  and  8  is  13 

6  and  8  is  14 

7  and  8  is  15 

8  and  8  is  16 

5  and  9  is  14 

6  and  9  is  15 

7  and  9  is  16 

8  and  9  is  17 

5  and  10  is  15 

6  and  10  is  16 

7  and  10  is  17 

8  and  10  is  18 

5  and  11  is  16 

6  and  11  is  17 

7  and  11  is  18 

8  and  11  is  19 

5  and  12  is  17 

6  and  12  is  18 

7  and  12  is  19 

8  and  12  is  20 

9  and  1  is  10 

10  and  1  is  11 

Hand  1  is  12 

12  and  1  is  13 

9  and  2  is  11 

10  and  2  is  12 

11  and  2  is  13 

12  and  2  is  14 

9  and  3  is  12 

10  and  3  is  13 

11  and  3  is  14 

12  and  3  is  15 

9  and  4  is  13 

10  and  4  is  14 

11  and  4  is  15 

12  and  4  is  16 

9  and  5  is  14 

10  and  5  is  15 

11  and  5  is  16 

12  and  5  is  17 

9  and  6  is  15 

10  and  6  is  16 

11  and  6  is  17 

12  and  6  is  18 

9  and  7  is  16 

10  and  7  is  17 

11  and  7  is  18 

12  and  7  is  19 

9  and  8  is  17 

10  and  8  is  18 

11  and  8  is  19 

12  and  8  is  20 

9  and  9  is  18 

10  and  9  is  19 

11  and  9  is  20 

12  and  9  is  21 

9  and  10  is  19 

10  and  10  is  20 

11  and  10  is  21 

12  and  10  is  22 

9  and  1 1  is  20 

10  and  11  is  21 

11  and  11  is  22 

12  and  11  is  23 

9  and  12  is  21 

10  and  12  is  22 

11  and  12  is  23 

12  and  12  is  24 

Since  0  has  no  value,  the  sum  of  any  number  and  0  is  the  number  itself; 
thus,  19  and  0  is  19. 


§1  ADDITION  AND  SUBTRACTION  13 

from  1  to  12;  it  should  be  thoroughly  committed  to  memory,  and 
to  do  this  the  reader  should  propose  to  himself  all  kinds  of  com- 
binations of  two  numbers  until  as  soon  as  the  numbers  are  named 
he  can  give  the  sum  almost  unconsciously.  When  at  work,  when 
walking,  or  at  any  time  when  he  has  leisure  he  should  practice  by 
saying  8  and  9  is  17,  9  and  5  is  14,  5  and  11  is  16,  9  and  8  is  17, 
etc.  until  he  can  name  the  sum  instantly  and  correctly. 

34.  To  Add  a  Digit  and  a  Number  Greater  than  12. — The  sum 
of  a  single  digit  and  a  number  greater  than  12  can  always  be 
obtained  mentally.  For  example,  consider  the  numbers  58  and 
9;  write  these  numbers,  one  above  the  other,  so  that  the  unit 
figures  stand  in  the  same  column,  and  draw  a  line  under  them. 

58  or  9 

9  58 

Now  add  the  unit  figures,  the  sum  being  17,  and  write  the  sum 
underneath  the  line  with  the  unit  figure  7  standing  in  the  same 
column  as  the  unit  figures  8  and  9.  Now  remembering  that  the  5 
in  58  represents  50  add  this  to  the  sum  of  the  unit  figures,  obtain- 
ing 67  for  the  sum  of  58  and  9 

(a)  (6)  (c)  (d) 

58  9  58  9 

9  58  J)  58 

17  17  67  67 

50  50 

67  07 

Now  notice  that  the  same  result  may  be  obtained  with  less 
figures,  in  fact  mentally,  from  the  following  considerations:  The 
sum  of  any  two  digits  cannot  exceed  18,  since  the  largest  digit  is 
9,  and  9  +  9  =  18;  18  =  10  +  8  or  1  ten  and  8  units.  Hence, 
when  two  numbers  are  added  as  above,  and  the  sum  of  the  unit 
figures  is  10  or  a  greater  number,  write  the  unit  figure  of  the  sum 
of  the  unit  figures  and  carry  the  1  ten  to  t  he  next  left-hand  column 
and  add  it  mentally  to  the  figure  or  figures  in  that  column.  Thus, 
in  (c)  above,  9  and  8  is  17;  write  7  and  carry  1,  which  added  to 
5  makes  6,  and  58  +  9  =  67.  In  (d),  8  and  9  is  17;  write  7  and 
carry  1,  which  added  to  5  makes  6. 

35.  The  student  will  find  it  greatly  to  his  advantage  to  practice 
adding  single  digits  to  numbers  less  than  100  at  every  convenient 
opportunity;  let  him  say,  for  example,  47  and  8  is  55,  23  and  9  is 
32,  36  and  4  is  40,  etc.,  the  entire  operation  being  performed 


14  ARITHMETIC  §1 

mentally.  As  soon  as  proficiency  has  been  obtained  in  adding 
simple  numbers  like  these,  there  will  be  little  trouble  in  adding 
any  number  of  figures  correctly.  Note  that  if  the  sum  of  the 
figures  in  the  right-hand,  or  lower,  order  does  not  exceed  9,  the 
digit  in  the  next  order  does  not  change;  but  if  the  sum  is  10  or 
more,  the  digit  is  increased  by  1. 

36.  If  the  larger  number  contains  more  than  two  figures,  the 
operation  is  exactly  the  same,  except  that  if  the  digit  in  the  next 
to  the  lowest  order  is  9,  the  digit  in  the  next  higher  order  must 
also  be  increased  by  1.  Thus  246  and  5  is  251,  since  5  +  6  =  11, 
and  4  +  1  =  5;  594  and  9  is  603,  since  9  and  4  is  13,  1  and  9  is  10, 
and  1  and  5  is  6;  1997  and  8  is  2005,  since  8  and  7  is  15,  1  and  9 
is  10,  1  and  9  is  10,  1  and  1  is  2;  etc. 

37.  To  Add  any  Two  Numbers. — The  sum  of  any  two  numbers 
is  found  as  follows:  Suppose  the  numbers  are  57,962  and  9,458. 
Write  them  so  that  the  figures  of  the  same  order  stand  in  the  same 
column  and  draw  a  line  underneath. 

57962 

9458 

67420 

Add  the  figures  in  the  right-hand  column  first,  obtaining  10;  write 
the  0  and  carry  the  1.  Add  the  1  to  the  5  in  the  next  column 
(making  6),  and  then  add  this  result  to  the  6  above  it,  obtaining 
12;  write  the  2  and  carry  the  1.  Add  1  to  the  4  in  the  next  column 
(making  5),  and  add  this  sum  to  the  9  above  it,  obtaining  14; 
write  the  4  and  carry  the  1.  Add  1  to  the  9  in  the  next  column 
(making  10),  and  add  this  sum  to  7  above  it,  obtaining  17;  write 
the  7  and  carry  the  1.  The  1  added  to  the  5  in  the  next  column 
gives  6,  which  is  written  as  shown. 

38.  The  procedure  is  exactly  the  same  when  either  or  both  the 
numbers  contain  decimals.  In  this  case  the  easiest  way  to  get  the 
figures  of  the  same  order  under  each  other  is  to  place  the  decimal 
points  under  each  other  in  the  same  column. 

26.943  0.434  87.51 

5.71  752.933  29.492 

32.653  753.367  117.002 

In  all  three  of  the  cases,  it  will  be  noted  that  units  are  placed 
under  units,  tens  under  tens,  tenths  under  tenths,  etc.,  and  that 


§1  ADDITION  AND  SUBTRACTION  15 

the  decimal  point  in  the  result  is  also  directly  under  the  decimal 
points  in  the  numbers  added. 

39.  To  Add  more  than  Two  Numbers. — 
Example. — Find  the  sum  of  8,  4,  6,  9,  7  and  3. 
Solution. —  8 

4 
6 
9 
7 

_3 

37  .4  ns. 
Explanation. — Beginning  at  the  bottom  of  the  column,  add  the  first 
two  numbers;  to  the  sum,  add  the  third  number;  to  this  sum,  add  the  fourth 
number;  etc.  Thus,  3  and  7  is  10;  10  and  9  is  19;  19  and  6  is  25;  25  and  4  is 
29;  29  and  8  is  37.  After  practicing  addition  in  this  manner,  until  a  certain 
degree  of  proficiency  has  been  attained,  the  reader  should  abbreviate  the 
process  by  naming  only  the  sums;  thus,  3,  10,  19,  25,  29,  37.  This  will 
greatly  increase  his  speed  in  adding. 


EXAMPLES 

Find  the  sum  of  the  following  sets  of  numbers. 

(1)3+5  +  1+9  +  7  =  ?  Ans.  25. 

(2)  4+2+6+8  =  ?  Ans.  20. 

(3)  3+4+6+8+7  =  ?  Ans.  28. 

(4)  5+8  +  1+6+9+4  =  ?  Ans.  33. 


40.  When  the  numbers  contain  more  than  one.  figure  use  the 
following  rule: 

Rule. — I.  Write  the  numbers  so  that  the  figures  of  the  same  order 
will  form  columns  having  units  under  units,  tens  under  tens,  tenth 
under  tenths,  etc.,  and  draw  a  line  under  the  bottom  row  of  figures. 

II.  Beginning  with  the  right-hand  column,  add  the  digits  in  that 
column,  and  write  the  unit  figure  of  the  sum  under  the  line  and  in  the 
column  that  was  added.  Add  the  tens  figure  of  the  sum  to  the  digits 
in  the  next  column  to  the  left,  add  (his  column  and  write  the  unit 
figure  of  the  sum  under  the  line  in  the  column  added.  Add  the  tens 
figure  of  the  sum  (if  any)  to  the  digits  in  the  third  column,  proceeding 
in  this  manner  until  the  addition  is  finished,  which  will  be  when  all 
the  columns  have  been  added. 

III.  7/  a  cipher  (0)  occurs  anywhere,  disregard  it,  since  it  does 
not  affect  the  sum. 


16  ARITHMETIC  §1 

Adding  the  tons  figure  of  the  sum  of  the  digits  in  any  column  to 
the  digits  in  the  next  column  to  the  left  is  called  carrying. 
Example.— 236  +  109  +  871  +  52  +  467  +  696  =  ? 
Solution. —  236 

109 
871 
52 
467 
696 
2431     Ans. 
Explanation*. — .Arrange  the  numbers  as  shown,  with  units  under  units, 
tens  under  tens,  etc.     Beginning  with  the  right-hand  column,  say  6,  13,  15, 
16,  25,  31;  write  the  1  under  the  line  and  in  the  column  just  added,  and 
carry  3  to  the  next  column.     Say  3,  12,  18,  23,  30,  33;  write  3  and  carry  3 
to  add  to  the  digits  in  the  next  column.    Say  3,  9,  13,  21,  22,  24.     As  there 
are  no  more  columns  to  add,  write  the  last  sum,  24,  as  shown.     The  sum  of 
all  the  numbers  is  2431. 

It  is  always  a  good  plan  to  write  the  abbreviation  Ans.  (which 
means  answer)  after  the  final  result  has  been  obtained. 

41.  If  some  of  the  numbers  contain  decimals,  they  are  added 
according  to  the  same  rule.  Arranging  the  numbers  with  units 
under  units,  etc.,  brings  the  decimal  points  under  one  another 
so  that  they  all  stand  in  the  same  column;  hence,  I  of  the  rule  in 
Art.  40  might  be  changed  to  read:  arrange  the  numbers  so  that 
the  decimal  points  stand  in  the  same  column.  The  rest  of  the 
rule  requires  no  change. 

Example.— Add  the  following  numbers:  403.7819,  21.875,  5.2742,  369.92, 
and  7923.917. 

Solution.—  403 .  7S19 

21.875 
5  2742 
369.92 
7923.917 
8724  70S1     Ans. 
Explanation. — Arranging  the  numbers  so  that  the  decimal  points  stand 
in  the  same  column,  say  2,  11,  and  write  1  and  carry  1.     Say  1,  8,  12,  17,  IS; 
write  8  and  carry  1.     Say  1.  2,  4,  11,  18,  26;  write  6  and  carry  2.     Say  2,  11, 
20.  22,  30,  37;  write  7  and  carry  3,  also  writing  the  decimal  point  before  the 
7.     Say  3,  6,  15,  20.  21.  24;  write  4  and  carry  2.     Say  2,  4,  10.  12;  write  2 
and  carry  1.     Say  1,  10.  13,  17;  write  7  and  carry  1.     Say  1,  8,  and  write  8. 
The  entire  sum  is  8724.7681. 

42.  In  bookkeeping,  and  when  making  out  bills,  business  state- 
ments, etc.,  the  decimal  point  is  usually  omitted,  a  vertical  line 
being  used  in  its  place.     This  practice  tends  to  prevent  mistakes 


§1  ADDITION  AND  SUBTRACTION  17 

and  saves  the  writing  of  decimal  points,  the  vertical  line  sepa- 
rating the  dollars  from  the  cents. 

Example.— Add  $06.72,  $243.27,  $127.85,  $37.40,  and  $101.32. 
Solution. —  $  60  72 

243  27 
12785 
37!40 
101  32 
$57050  Ana. 

Explanation. — Here  the  vertical  line  is  used  instead  of  the  decimal 
points.  The  addition  is  performed  in  the  usual  manner,  and  the  sum  is 
found  to  be  $576.56. 

43.  Checking  Results. — In  business  and  in  engineering,  it  is  of 
the  utmost  importance  that  the  final  result  be  correct.  While 
mistakes  or  blunders  are  liable  to  occur,  even  when  the  greatest  care 
is  taken,  they  must  be  overcome  by  some  method  of  detecting 
them,  and  corrections  must  be  made  before  the  final  result  is 
accepted.  In  checking  the  work  of  addition,  the  various  num- 
bers may  be  added  again,  but  if  this  is  done  immediately  after  the 
first  adding,  the  same  mistake  is  likely  to  be  made  again.  A 
better  way  is  to  add  the  several  columns  downward;  this  causes 
a  change  in  the  sequence  of  the  digits  added,  and  tends  to  prevent 
a  second  mistake  like  the  first.  If  the  same  result  is  obtained 
when  adding  down  as  was  obtained  when  adding  up,  the  work  is 
presumed  to  be  correct;  but  if  a  different  result  is  obtained,  repeat 
the  work  until  the  same  result  is  obtained  by  adding  both  ways. 
This  practice  of  testing  the  work  to  see  if  it  is  correct  is  called 
checking,  and  the  method  used  in  checking  is  called  a  check.  The 
reader  should  apply  the  check  by  adding  down  to  the  two  examples 
of  the  preceding  article. 


EXAMPLES 

(1)  26.48  4-  360.72  +  54.008  +  7.205  +  509.045  =  ?  Ans.  957.518. 

(2)  19081  +  89320  +  20358  +  33140  +  7608  =  ?  At  <.    170013. 

(3)  5818  4-  8720  +  0791  +  9809  +  12463       762  =  ?  Ans.    1 : 

(4)  0.317  +  49.42  +  17.718  +  5.801  +  S3.77  =  ?  Ans.   163.026. 

(5)  The  weights  of  seven  bags  <>f  alum  air  292  pounds,  306  pounds,  301 
pounds,  298  pounds,  297  pounds,  :;<>7  pounds,  ami  :5<):{  pounds;  what  is  their 
total  weight?  Ana.  2103  pounds. 

(0)  What  is  the  weight  of  six  mils  of  paper,  if  the  rolls  weigh  967  pounds, 
1048  pounds,    1075  pounds,   993  pounds,   986  pounds,   and   979  pounds? 

Aim.  0038  pounds. 


18  ARITHMETIC  §1 

(7)  Paid  the  following  amounts  for  bales  of  rags:  $18.82,  $18.94,  $19.11, 
$19.20,  $1S.S5,  $18.97.     What  was  the  total  amount  paid?      Ana.  $113.89. 

(8)  Six  barrels  of  rosin  were  placed  on  an  elevator;  if  the  barrels  weighed 
389  pounds,  411  pounds.  395  pounds,  399  pounds,  408  pounds,  and  406 
pounds,  how  much  must  the  elevator  raise,  if  it  carries  in  addition  to  the 
rosin  a  man  weighing  168  pounds?  Ans.  2576  pounds. 


SUBTRACTION 

44.  Subtraction  means  taking  away.  In  arithmetic,  subtrac- 
tion is  the  process  of  taking  one  number  from  another  number,  or 
it  is  the  process  of  finding  how  much  greater  one  number  is  than 
another.  If  a  person  has  10  cents  and  spends  4  cents,  he  will 
have  6  cents  left.  Here  4  cents  are  taken  from  10  cents,  and  6 
cents  are  left.  The  arithmetical  operation  of  taking  4  cents 
from  10  cents  is  called  subtraction.  Again,  how  much  greater  is 
10  cents  than  4  cents'?  If  6  cents  are  added  to  4  cents,  the  sum 
is  10  cents;  hence,  10  cents  is  6  cents  greater  than  4  cents.  The 
same  result  will  be  obtained  by  subtracting  4  cents  from  10  cents. 

45.  The  sign  of  subtraction  is  — ,  a  short  horizontal  line;  it  is 
read  minus,  and  means  less;  it  indicates  that  the  number  following 
it  is  to  be  subtracted  from  the  number  preceding  it.  Thus,  10  — 
4  =  6;  here  4  is  to  be  subtracted  from  10,  the  result  being  6;  the 
expression  may  be  read  either  as  10  minus  4  equals  6  or  as  10  less 
4  equals  6.  The  longer  expression,  4  subtracted  from  10  equals 
6,  is  also  correct. 

46.  The  number  subtracted  is  called  the  subtrahend,  and  the 
number  from  which  the  subtrahend  is  subtracted  is  called  the 
minuend;  the  result  is  called  the  remainder  or  difference.  When 
it  is  not  desired  to  be  specific  regarding  the  order  of  the  numbers, 
the  word  difference  is  generally  used;  thus,  the  difference  between 
4  and  10  or  the  difference  between  10  and  4  is  6.  But  when  the 
order  of  the  numbers  is  definitely  stated,  as  10  minus  4  is  6,  6 
is  called  the  remainder,  it  being  what  is  left  after  taking  4  from  10. 
In  an  expression  like  15  -  9  =  6,  15  is  the  minuend,  9  is  the 
subtrahend,  and  6  is  the  remainder. 

Before  explaining  the  general  process  of  subtraction,  the  two 
following  principles  or  laws  should  be  thoroughly  understood: 

47.  Principle  I. — If  the  same  number  be  added  to  both  minuend 
and  subtrahend,  the  remainder  is  unchanged.  For  example,  15  — 
9  =  6;  if  now,  10  be  added  to  both  15  and  9,  they  become  25 


§1  ADDITION  AND  SUBTRACTION  19 

and  19  respectively,  and  25  —  19  =  6,  the  same  remainder  as 
before.  The  reason  is  evident;  the  10  added  to  the  subtrahend  is 
subtracted  from  the  10  added  to  the  minuend,  thus  leaving  the 
minuend  and  subtrahend  the  same  as  before  the  10  was  added. 
A  like  result  will  be  obtained,  no  matter  what  number  is  added; 
thus,  adding  7  to  15  and  9,  these  numbers  become  22  and  16,  and 
22  —  16  =  6,  as  before. 

48.  Principle  II. — //  the  remainder  be  added  to  the  subtrahend, 
the  sum  will  be  the  minuend.  For  instance,  if  a  stick  is  13  feet 
long  and  5  feet  are  cut  off,  the  Length  of  the  stick  is  then  8  feet. 
Here  13  -  5  =  8,  and  5  is  the  subtrahend  and  8  is  the  remainder. 
But  when  the  two  parts  of  the  stick  are  placed  together,  they 
must  be  equal  in  length  to  the  original  stick;  that  is,  5  -f  8  =  13, 
or  subtrahend  +  remainder  =  minuend.  Hence,  instead  of 
saying  5  from  13  leaves  8,  it  will  be  equally  proper  to  say  "what 
number  added  to  5  will  make  13?"  this  number  is  8;  therefore, 
13  -  5  =  8,  because  5  +  8  =  13. 

49.  The  reader  will  find  it  greatly  to  his  advantage  to  reverse 
the  addition  table  of  Art.  33.  Thus,  9  and  6  is  15;  15  less  9  is  6, 
and  15  less  6  is  9.  Again,  8  and  5  is  13;  13  less  8  is  5,  and  13  less 
5  is  8.  He  should  practice  in  this  manner  both  addition  and 
subtraction  in  spare  moments,  until  the  result  presents  itself 
instantly  as  soon  as  the  numbers  are  named. 

50.  If  the  right-hand  figure  of  the  minuend  is  smaller  than  the 
digit  to  be  subtracted,  add  10  to  it,  subtract,  and  then  subtract  1 
from  the  next  left-hand  figure  of  the  minuend;  thus,  34  -  8  = 
26.  Here  10  is  added  to  4,  making  14;  8  from  14  leaves  6;  then 
subtracting  1  from  3,  the  remainder  is  2.  This  is  in  accordance 
with  Principle  I,  Art.  47,  since  10  is  added  to  the  minuend  (really 
making  it  44,  but  expressed  as  30  +  14)  and  10  is  added  to  the 
subtrahend  (really  making  it  18).  The  result  is  correct;  since, 
by  Principle  II,  8  +  26  =  34;  also  18  +  26  =  44.  It  is  thus 
seen  that  36  -  9  =  27,  43  -  7  =  36,  81  -  4  =  77,  etc.  The 
reader  should  practice  these  combinations  also. 

61.  Bearing  the  foregoing  in  mind,  the  following  La  the  rule  for 
subtraction : 

Rule  I. — Place  the  numbers  as  in  addition,  with  the  subtrahend 
under  the  minuend,  and  with  units  under  units,  tens  under  tens, 
etc.  or  with  the  decimal  points  in  the  same  column. 

II.  Beginning  with  the  right-hand  digit  of  the  subtrahend,  sub- 


20  ARITHMETIC  §1 

tract  it  from  the  digit  above  it  in  the  minuend;  do  the  same  with  the 
next  digit  or  figure  to  the  left,  proceeding  in  this  manner  until  all 
the  figures  in  the  subtrahend  have  been  subtracted  from  the  figures 
above  them  in  the  minuend. 

III.  If  a  figure  in  the  minuend  is  a  cipher  or  is  smaller  in  value 
than  the  figure  under  it  in  the  subtrahend,  add  10  to  it  before  sub- 
tracting, and  then  carry  1  to  the  next  figure  of  the  subtrahend  before 
subtracting  it  from  the  figure  above. 

IV.  Having  found  the  remainder,  add  it  to  the  subtrahend,  figure 
by  figure,  and  if  the  sum  equals  the  minuend,  the  remainder  is  very 
probably  correct.     This  last  operation  is  a  check. 

Example  1.— From  93S75  take  72032. 

Solution. —  93875 

72032 

21843     Ana. 

Explanation. — Writing  the  numbers  as  directed  in  I  of  the  rule,  begin 
with  the  right-hand  digit  and  say  2  from  5  is  3,  3  from  7  is  4,  write  8  (since  0 
from  any  number  leaves  the  number),  2  from  3  is  1,  and  7  from  9  is  2.  The 
remainder,  therefore,  is  21.S33.  To  check  this  result,  add  the  remainder 
to  the  subtrahend;  thus.  3  and  2  is  5.  -4  and  3  is  7,  8,  1  and  2  is  3,  and  2  and 
7  is  9.  The  sum  being  the  same  as  the  minuend,  the  work  is  very  probably 
correct. 

Example  2.— From  $1437.50  take  $918.27. 

Solution.—  $1437  50 

918.27 
$  519.23     Ans. 

Explanation. — Since  7  cannot  be  subtracted  from  0,  add  10  to  0.  obtain- 
ing 10;  then  7  from  10  is  3.  Carry  1  to  2  making  it  3,  and  3  from  5  is  2. 
Write  the  decimal  point.  Then,  since  8  cannot  be  taken  from  7,  add  10, 
making  17;  and  8  from  17  is  9.  Carry  1  to  1  making  it  2,  and  2  from  3  is  1. 
Lastly,  9  from  14  is  5.  If  the  remainder  be  added  to  the  subtrahend,  the 
sum  is  the  minuend.     Hence,  the  correct  remainder  is  $519.23. 

Example  3. — There  are  four  piles  of  papers;  the  first  pile  contains  986 
sheet?,  the  second  pile  contains  753  sheets,  the  third  pile  contains  875  sheets, 
and  the  fourth  pile  contains  103S  sheets;  they  were  all  placed  in  one  pile, 
and  2369  sheets  were  taken  away;  how  many  sheets  were  left? 

Solution. — The  first  step  is  to  find  how  many  sheets 
there  were  all  together.     This  found  by  adding  the  ^86  sheets 

number  of  sheets  in  each  pile,  the  sum  or  total  being  ,0c* 

3652  sheets.     The  next  step  is  to  subtract  from  this  *■" 

sum  the  number  of  sheets  that  were  taken  away,  the         1038 
remainder  thus  found  being  1283  sheets.     To  check,         3652  sheets 
add  down,  then  add  the  remainder  to  the  subtrahend,         2369 
obtaining  3652,  the  minuend.     Since  the  work  checks         1283  sheets  Ans. 
in  both  cases,  it  is  probably  correct. 


§1  ADDITION  AND  SUBTRACTION  21 

Example  4.— From  10.125  take  9.3125. 
Solution.—  19.1250 

9  3125 

9.8126    Am. 

Explanation. — There  is  no  figure  in  the  minuend  above  the  right-hand 
figure  of  the  subtrahend;  but  since  the  subtrahend  contains  a  decimal, 
it  is  allowable  (for  reasons  that  will  be  explained  Inter)  to  annex  ciphers 
to  the  minuend  until  it  contains  the  same  number  of  figures  in  the  decimal 
part  that  there  are  in  the  subtrahend.  The  subtraction  is  then  performed 
in  the  regular  manner.  Should  there  be  more  figures  in  the  decimal  part  of 
the  minuend  than  in  the  decimal  part  of  the  subtrahend,  annex  ciphers  to 
the  subtrahend  until  the  decimal  parts  of  the  subtrahend  and  minuend  both 
contain  the  same  number  of  figures,  before  subtracting. 

52.  Although  subtraction  is  a  much  simpler  and  easier  opera- 
tion than  addition,  it  is  probable  that  more  mistakes  are  made  in 
subtraction  than  in  addition,  the  reason  being  that  the  operator 
fails  to  check  by  "adding  back"  and  is  more  inclined  to  be  care- 
less. It  is  well  not  to  attempt  to  work  too  fast  at  first;  try  to 
secure  accuracy,  and  speed  will  come  with  practice.  This  advice 
applies  to  every  process  of  computation. 


EXAMPLES 

(1)  From  417.23  take  273.58.  Arts.  143.65. 

(2)  From  54.375  take  4S.G5G25.  Am.  5.71875. 

(3)  From  484814  take  258177.  Ans.  226637. 

(4)  From  1002070  take  304098.  Ans.  697972. 

(5)  2033.4238  -  1792.3917  =  ?  Ans.  241.0321. 

(6)  From  a  pile  of  pulp  weighing  14,253  tons  9,468  tons  were  sold;  how 
much  pulp  remained  in  the  pile?  Am    4,7S5  tons. 

(7)  Three  cars  were  received  containing  58,024  pounds  of  sulphur;  when 
the  stock  was  cleaned  up,  mill  figures  showed  that  55,638  pounds  had  been 
used;  how  much  had  been  lost  in  handling?  Ans.  2,386  pounds 

(8)  To  a  pile  of  pulp  weighing  4,826  tons,  527  tons  were  added;  how  much 
was  left  after  3,962  tons  had  been  sold?  Ana.  1,391  tons. 

(9)  A  bleach  liquor  tank  contained  925  gallons;  how  much  renamed  after 
146  gallons  had  been  used?  A^.  771)  gallons. 


22  ARITHMETIC  §1 

MULTIPLICATION  AND  DIVISION 


MULTIPLICATION 

53.  In  arithmetic,  multiplication  may  be  defined  as  a  short 
method  of  adding  (or  finding  the  sum  of)  several  equal  numbers. 

For  instance,  suppose  there  are  6  books, 
(Q)  each  book  containing  1892  pages;  how 

1892PageS  many  pages   are  in  the   6   books?     To 

18g2  find   the   total    number,    1892   may   be 

1892  written  six  times  as  shown  at  (a),  and  the 

1892  sum  found,  which  is  11352  pages.     By 

1892  using  the  process  of  multiplication,  how- 

11352  pages    Ans.  ever,  the  work  will  be  arranged  as  shown 

.  at  (b),  and  1892  is  then  multiplied  by  6, 

18Q2  the   result   being   the   same   as    before. 

6  Note  the  great  saving  in  figures  even  in 

11352  pages    Ans.  this    simple    case.     Suppose    there    had 

been,   say,    576   books   each   containing 

1892  pages;  it  would  then  be  practically  impossible  to  find  the 

total  number  of  pages  by  the  method  shown  at  (a),  but  the  total 

can  easily  be  found  by  multiplication. 

54.  The  number  multiplied  (1892  in  Art.  53)  is  called  the  multi- 
plicand; the  number  used  to  multiply  the  multiplicand  (and 
which  shows  how  many  times  the  multiplicand  is  to  be  added)  is 
called  the  multiplier  (this  is  6  in  Art.  53) ;  the  result  of  the  opera- 
tion of  multiplication  is  called  the  product  (in  Art.  53,  11352  is 
the  product). 

When  it  is  not  desired  to  be  specific  and  make  a  special  dis- 
tinction between  the  multiplicand  and  multiplier,  these  two  num- 
bers are  called  factors;  in  Art.  53, 1892  and  6  are  factors  of  11352, 
their  product. 

55.  The  sign  of  multiplication  is  an  oblique  (St.  Andrew's) 
cross,  which  is  written  between  the  factors;  it  is  read  times  or 
multiplied  by.  Thus,  6  X  8  =  48  is  read  either  as  six  times 
eight  equals  forty-eight  or  as  six  multiplied  by  eight  equals  forty- 
eight;  here  6  and  8  are  factors,  6  being  the  multiplier  in  the  first 
case  and  8  being  the  multiplier  in  the  second  case. 

56.  Insofar  as  the  product  is  concerned,  it  makes  no  difference 
which  of  two  factors  is  considered  as  the  multiplier,  since  6  times 


§1 


MULTIPLICATION  AND  DIVISION 


23 


8  is  the  same,  for  instance,  as  8  limes  6,  (lie  product  in  both  cases 
being  48;  this  may  be  expressed  mathematically  as  6  X  8  =  8 
X  6  =  48. 

57.  Before  one  can  multiply,  it  is  necessary  that  he  memorise 
the  multiplication  table.  This  may  take  a  little  time,  but  it  is 
absolutely  necessary  if  the  reader  is  to  be  successful  in  this 
subject. 

MULTIPLICATION  TABLE 


1  tirr.es 
1  times 
1  times 
1  times 
1  times 
1  times 
1  times 
1  times 
1  times 
1  times  10  is 
1  times  11  is 
1  times  12  is 


1  is 

2  is 

3  is 

4  is 

5  is 

6  is 

7  is 

8  is 

9  is 


2  times 
2  times 
2  times 
2  times 
2  times 
2  times 
2  times 
2  times 
2  times 
2  times 
2  times 
2  times 


1  is 

2  is 

3  is 

4  is 

5  is 

6  is 

7  is 

8  is 

9  is 

10  is 

11  is 

12  is 


mea 
mea 
mea 

mis 

mea 

IlieS 

mea 
mea 
mea 

II1C.4 

mea 
mea 


1  is 

2  is 

3  is 

4  is 

5  is 

6  is 

7  is 

8  is 

9  is 

10  is 

11  is 

12  is 


4  t 

■I  t 

4  t 

4  t 

4  t 

4  t 

4  t: 

1  t 

4  t 

4  t 

4  t 

4  t 


Ml  OS 
111  OS 

mrs 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 


1  is 

2  is 

3  is 

4  is 

5  is 

6  is 

7  is 
7  is 
9  is 

10  is 

11  is 

12  is 


6  t 

5  t 

5  t 

5  t 

5  t 

5  t 

5  t 

5  t 

5  t 

5  t: 

5  t 

5  t 


mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 


1  is  5 

2  is  10 

3  is  15 

4  is  20 

5  is  25 

6  is  30 

7  is  35 

8  is  40 

9  is  45 

10  is  50 

11  is  55 

12  is  60 


6  t: 

6  t 

6  t 

6  t: 

6  t 

6  t 

6  ti 

6  t 

6  t 

6  ti 

6  ti 

G  t 


mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mes 
mea 
mea 
mea 


1  is 

2  is 

3  is 

4  is 

5  is 

6  is 

7  is 

8  is 

9  is 

10  is 

11  is 

12  is 


7  t 

7  t: 

7  t 

7  t 

7  t 

7  t 

7  t 

7  t 

7  t 

7  t 

7  t 

7  t 


mes 
mes 
mes 
mes 
mea 
mes 
mes 
mes 

mea 

mes 
mes 


1  is  7 

2  is  14 

3  is  21 

4  is  28 

5  ib  35 

6  is  42 

7  is  49 

8  is  56 

9  is  63 

10  is  70 

11  is  77 

12  is  84 


8  times 
8  times 
8  times 
8  times 
8  times 
8  timea 
8  times 
8  times 
8  times 
8  times 
8  times 
8  times 


1  is 

2  is 

3  is 

4  is 

5  is 

6  is 

7  is 

8  is 

9  is 

10  is 

11  is 

12  is 


9  t 

9  t 

9  ti 

9  t 

9  t 

9  t 

9  t 

9  t 

9  t 

9  t 

g  t 

9  t 


imes 

1 

s 

9 

10  times 

1  i 

imes 

2 

s 

18 

10  times 

2  u 

imes 

3 

s 

27 

10  times 

3  ii 

imes 

4 

s 

36 

10  times 

4  it 

-imes 

5 

s 

45 

10  times 

5  ii 

imes 

6 

B 

54 

10  times 

6i 

imes 

7 

s 

63 

10  times 

7  i 

,imes 

8 

a 

72 

10  times 

8  ii 

imes 

9 

s 

81 

10  times 

9  i 

imes 

10 

a 

90 

10  times 

10  i 

-imes 

11 

0 

99 

10  times 

11  i 

imes 

12 

8 

108 

10  times 

12  i 

8      10 

s  20 
s  30 
s    40 

50 
s  60 
a  70 
s    80 

90 
s  100 
s  110 
s  120 


1 1  timea 
1 1  timea 
11  times 
1 1  timea 
1 1  timea 
1 1  times 
1 1  times 
1 1  timea 
1 1  timea 

1  1  times 
1  1  times 
1  1   times 


11 

33 
33 
s  44 
s  55 
s  66 
s  77 
s  88 
s  99 
s  110 
s  121 
s  132 


13  t 

12  t 
12  t 
12  t 
12  t 
12  t 
12  t 
12  t 
12  t 
12  t 

12  t 

13  t 


imes 

1  i 

imes 

2  i 

imes 

3  i 

imes 

4  i 

imes 

5  i 

imes 

6  i 

imes 

7  i 

imes 

8  i 

imes 

9  i 

imes 

10  i 

imee 

11  i 

imes 

12  i 

13 
M 
86 

48 
60 

72 
-1 
96 

a  108 

s  120 
s  132 
s  144 


The  product  of  any  number  and  0  is  0  (9  X  0  =  0,  167  X  0  = 
0,  etc.);  this  is  evident,  since  the  sum  of  any  number  of  0's  cannot 
make  an  integer. 

It  will  be  noted  by  reference  to  the  table  that  the  product  of  any 
number  and  1  is  the  number  itself;  thus,  409  X  1  =  409,  since 


24  ARITHMETIC  §1 

409  X  1  is  409  l's,  which  is,  of  course,  409.  Note,  also,  that  the 
product  of  any  number  and  10  is  the  number  itself  with  a  cipher 
(0)  added  at  the  right;  thus,  7  X  10  =  70,  526  X  10  =  5260, 
etc.  Note,  again,  that  the  product  of  11  and  any  number  not 
greater  than  9  is  the  number  repeated;  thus,  3  X  11  =  33,  6  X 
11  =  66,  9  X  11  =  99,  etc. 

.  The  reader  should  repeat  the  different  parts  of  the  table  to 
himself  at  odd  times  until  it  becomes  so  firmly  impressed  on  his 
memory  that  as  soon  as  any  two  numbers  are  named,  their  pro- 
duct will  instinctively  name  itself. 

58.  To  Multiply  any  Number  by  a  Single  Digit. — Multiply 

31415927  31415927   by   7.     Here  31415927  is  the 

7  multiplicand    and    7,    the    multiplier,  is 

2199114S9  Ana  written  under  the  right-hand  figure  of  the 
multiplicand.  Draw  a  line  under  the 
factors,  as  shown,  and  multiply  the  right-hand  figure  of  the 
multiplicand  by  7  the  multiplier,  obtaining  7  X  7  =  49.  Write 
the  unit  figure  of  the  product  under  the  unit  figure  of  the  multipli- 
cand, and  carry  4,  the  tens  figure  of  this  product.  Then  say  7 
times  2  is  14,  to  which  add  4,  the  figure  carried,  making  18;  write 
8  as  shown  and  carry  1.  Say  7  times  9  is  63,  add  the  1  carried, 
making  64;  write  4  and  carry  6.  Say  7  times  5  is  35  and  6  (the 
figure  carried)  is  41;  write  1  and  carry  4.  Say  7  times  1  is  7  and 
4  (the  figure  carried)  is  11;  write  1  and  carry  1.  Say  7  times  4  is 
28  and  1  is  29;  write  9  and  carry  2.  Say  7  times  1  is  7  and  2  is  9; 
write  9  and  there  is  nothing  to  carry.  Lastly,  say  7  times  3  is 
21,  which  write.  Every  figure  in  the  multiplicand  has  now  been 
multiplied  by  the  multiplier,  7,  and  the  product  is  219,911,489. 
Had  the  multiplier  been  70,  700,  7000,  etc.,  the  process  would 
have  been  exactly  the  same,  except  that  after  the  product  was 
found,  as  many  ciphers  would  have  been  annexed  to  the  product 
as  there  were  ciphers  to  the  right  of  the  right-hand  digit  of  the 
multiplier;  thus,  31,415,927  X  7000  = 
15927000  219,911,489,000.     The  work  is  shown  in 

219911489000  An*.  the    marSin-     First    multiply    by    7    as 

before;  then  to  the  product,  annex  three 
ciphers,  because  there  are  three  ciphers  to  the  right  of  the  right- 
hand  digit  of  the  multiplier . 

59.  To  Multiply  When  the  Multiplier  Contains  Two  or  More 
Digits. — Place  the  multiplier  under  the  multiplicand,  with  the 


§1  MULTIPLICATION   AND  DIVISION  25 

right-hand  digit  of  the  multiplier  under  the  right-hand  digit  of 
the  multiplicand.     Multiply  by  the  right-hand  digit  of  the  multi- 
plier and  write  the  product  figure  by  figure  under  the  multiplier, 
as  shown  in  the  margin.    This  result   is 
428095  called  the  first  partial  product.     Note  that 

37042  after  multiplying  9  by  2  there  is  1  to  carry; 

856190  then  say  2  limes  naught  is  naught  and  1  is 

onor!vrn°  1,  which  write  as  shown.     Now  multiply  by 

299b(>(>50  ,  .        .  . 

12842S5  the  next  figure  to  the  left  of  2,  m  tins  case 

15857494990  Am.  4.  Say  4  times  5  is  20;  write  the  cipher 
one  place  to  the  left  of  the  right-hand  figure 
of  the  first  partial  product,  thus  bringing  the  cipher  under  the 
figure  multiplied  by.  Continue  the  multiplication  by  -1,  obtain- 
ing 1712380,  which  is  called  the  second  partial  product.  The  third 
figure  to  be  used  as  a  multiplier  is  0,  and  since  any  number 
multiplied  by  0  is  0,  write  a  cipher  one  place  to  the  left  of  the 
right-hand  figure  of  the  second  partial  product,  which  brings  it 
directly  under  the  cipher  in  the  multiplier.  Now  multiply  by 
the  next  figure,  7,  of  the  multiplier.  Say  7  times  5  is  35,  write 
5  alongside  the  cipher  and  carry  3;  this  brings  the  5  under  the 
figure  used  as  a  multiplier,  and  makes  the  third  row  of  figures 
29966650,  the  third  partial  product,  which  is  equal  to  428095  X 
70.  Finally,  multiply  by  3,  the  left-hand  digit  of  t  be  mult  iplier, 
and  the  result  is  the  fourth  partial  product,  the  right-hand  figure 
of  which  is  written  under  3,  the  number  multiplied  by.  Now 
adding  the  four  partial  products,  the  sum  is  15,857,494,990, 
which  is  the  entire  product,  or  the  result  sought, 

If  there  are  ciphers  to  the  right  of  the  mult  iplicand  or  multiplier 
or  both,  pay  no  attention  to  them  until  after  the  product  has 
been  found  as  just  described.  Then  annex  to  the  entire  product 
as  many  ciphers  as  there  arc  ciphers  to  the  right  of  either  or  both 

factors.    For    instance,     to    multiply 

526700  526700  by  205000,  arrange  as  shown  in 

205000  the  margin,  with  the  right-hand  digits 

20335  of  the  multiplicand  and  multiplier  under 

105340       _     ,  each  other.     Multiply  5267  by  205.  the 

107973500000  Am.  ^^    ^  ^^  ^  ^   ,„.„ 

ciphers  to  the  right  of  one  factor  and  three  to  the  right  of  the 
other  factor;  hence,  annex  2  +  3  =  5  ciphers  to  the  right  of  the 
entire  product,  which  is  thus  found  to  be  107,973,500,000. 


26  ARITHMETIC  §1 

60.  Rule. — Write  the  multiplier  under  the  multiplicand  with  the 
right-hand  digits  under  each  other.  Beginning  with  the  right-hand 
digit  of  the  multiplier,  and  proceeding  to  the  left,  multiply  the  upper 
factor  by  each  figure  of  the  lower  factor,  or  multiplier,  writing  the 
right-hand  figure  of  each  partial  product  under  the  figure  used  as  a 
multiplier.  Then  add  the  partial  products,  and  the  sum  will  be  the 
entire  product. 

61.  Check  for  Multiplication. — The  best  way  to  check  multi- 
plication is  to  employ  the  process  called  "casting  out  nines." 
This  consists  in  dividing  (the  operation  of  dividing  will  be  con- 
sidered in  the  next  article)  the  two  factors  by  9,  multiplying  the 
remainders,  and  if  the  product  is  greater  than  9,  divide  that  by  9; 
note  the  remainder.  Then  divide  the  entire  product  by  9,  and 
if  the  remainder  is  the  same  as  that  first  obtained,  the  work  is 
very  probably  correct.  If  the  two  remainders  differ,  however, 
then  the  work  is  wrong,  some  mistake  having  been  made.  In- 
stead of  dividing  by  9.  the  remainder  may  be  found  by  adding  the 
digits;  if  the  sum  is  greater  than  10,  add  the  digits  of  the  sum,  pro- 
ceeding in  this  manner  until  a  single  digit  has  been  found,  which 
will  be  the  remainder  when  the  number  is  divided  by  9.  Thus, 
consider  the  number  7854;  the  sum  of  the  digits  is  7  +  8  +  5  +  4 
=  24 ;  2  +  4  =  6,  and  6  is  the  remainder  when  7854  is  divided  by  9. 

Applying  this  check  to  the  first  example  of  Art.  59,  the  sum  of 
the  digits  in  the  multiplicand  is  4  +  2  +  8+9  +  5  =  28, 2+8 
=  10,  and  1  +  0  =  1;  the  sum  of  the  digits  in  the  multiplier  is 
3  +  7  +  4  +  2  =  16,  and  1  +  6  =  7;  then  1X7  =  7.  The 
sum  of  the  digits  in  the  entire  product  is  1  +  5  +  8  +  5  +  7  +  4  +  9 
+  4  +  9  +  9  =  61,  and  6  +  1=7.  Since  the  remainders  are 
both  7,  the  work  is  very  probably  correct.  When  adding  the 
digits  in  this  manner,  it  is  not  necessary  to  add  any  9's;  thus,  in 
the  foregoing,  the  remainder  for  the  multiplicand  is  4  +  2  +  8  + 

5  =  19  or  1,  and  the  remainder  for  the  product  is  1  +  5  +  8  +  5 
+  7  +  4  +  4  =  34,  and  3  +  4  =  7,  both  results  being  the  same 
as  obtained  before. 

Applying  this  check  to  the  second  example  of  Art.  59,  5  +  2  + 

6  +  7  =  20,  or  2;  2  +  5  =  7;  2  X  7  =  14,  and  1  +  4  =  5; 
1  +  7  +  7  +  3  +  5  =  23,  and  2  +  3  =  5.  Since  the  remain- 
ders are  the  same,  the  work  is  very  probably  correct. 

The  reader  is  strongly  advised  to  apply  this  check  in  every 
case. 


§1  MULTIPLICATION  AND  DIVISION  27 

EXAMPLES 

(1)  7854  X  203S  =  ?  Ans.  16,006,452. 

(2)  230258  X  90057  =  ?  Ans.  20,736,344,706. 

(3)  31831  X  31416  =  ?  Ans.  1,000,002,696. 

(4)  543836  X  46S8  =  ?  Ans.  2,549,503,168. 

(5)  197527  X  98743  =  ?  Ans.   19,504,408,561. 

(6)  295369  X  700405  =  ?  Ans.  200,877,924,445. 

(7)  The  average  consumption  of  coal  by  a  mill  per  year  is  28,750  tons;  if 
the  average  cost  per  ton  is  $7.00,  what  is  the  annual  cost  to  the  mill  for  coal? 

Ans.  §201,250. 

(8)  During  a  certain  period,  a  mill  consumed  686  tons  of  alum ;  if  the  price 
paid  for  the  alum  was  $49.00  per  ton,  how  much  was  the  total  amount  paid 
for  the  alum?  A ns.  $33,614. 

(9)  What  is  the  value  of  13,908  cords  of  wood  at  $11.00  per  cord? 

Ans.  $152,988. 

(10)  How  much  must  be  paid  for  12  cans  of  dye  stuff,  if  each  can  weighs 
47  pounds  and  the  dye  stuff  is  worth  $19.00  per  pound?  Ans.  $10,716. 

(11)  At  different  times  a  certain  mill  sold  paper  as  follows:  27,848 
pounds  at  14  cents  per  pound;  17,005  pounds  at  18  cents  per  pound;  9,990 
pounds  at  19  cents  per  pound;  and  36,476  pounds  at  15  cents  per  pound. 
How  much  was  received  from  these  sales?  Ans.  $14,329.12. 


DIVISION 

62.  Division  means  a  partition,  a  separating  into  parts.     In 

arithmetic,  division  is  the  process  of  finding  how  many  times 

larger  one  number  is  than  another;  thus,  since  24 

11352  *s  ®  times  4,  24  is  4  times  as  large  as  6  or  6  times  as 

1892  large  as  4.     Division  may  also  be  defined  as  the 

9460  process   of   separating   a  number  into  a   required 

1892  number  of  equal  parts;  thus,  24  may  be  separated 

7568  into  6  equal  parts  of  4  each  or  4  equal  parts  of  6 each. 

1892  Just    as    multiplication    is   a   short    process   or 

5676  method  of  adding  equal  numbers  so  division  is  a 

1892  short  process  or  method  of  subtracting  continuously 

3784  until  the  remainder  is  0  or  less  than  the  subtra- 

1 S9'' 

— —  hend.     For  example,  referring  to  Art.  53,  suppose 

1RQ2  there  are  11,352  pages  in  a  certain  number  of  books 

oooo  eac*1  contamm8  1892  Patf('s-  aml  n  is  required  to 

find  the  number  of  books.     The  work  might  be 

^  „     done  as  shown  at   (a)  in  the  margin,  subtracting 

1892)1135?(G     1892  from  11352,  fchen  sul,,rar,m-  1S!'2  fn,m  tho 
—       remainder,     continuing     this     process     until     the 

remainder  becomes  0  or  less  than  the  subtrahend. 


28  ARITHMETIC  §1 

In  this  case,  the  remainder  is  0;  and  as  the  subtraction  was 
performed  6  times,  there  are  6  books.  By  the  process  of  divi- 
sion, as  shown  at  (b),  1892  is  contained  6  times  in  11352,  because 
1892  X  6  =  11352.  Note  the  great  saving  of  figures  in  the 
second  method.  Had  it  been  known  that  the  total  number  of 
pages  was  11,352,  the  number  of  books  was  6,  and  it  were 
required  to  find  the  number  of  pages  in  each  book,  it  would  be 
necessary  to  subtract  6,  by  the  method  shown  at  (a),  1892  times, 
which  is  practically  impossible. 

In  multiplication,  the  object  to  be  attained  is  to  find  the 
product  of  two  numbers  (factors);  in  division,  the  object  is  to 
divide  a  number  into  two  factors,  one  of  them  being  given. 

63.  The  number  that  is  to  be  divided  into  two  factors  is  called 
the  dividend;  the  given  factor,  which  is  divided  into  the  dividend, 
is  called  the  divisor;  the  other  factor,  which  is  obtained  by  divid- 
ing the  dividend  by  the  divisor,  is  called  the  quotient;  anything 
that  is  left  over  after  the  division  has  been  performed  is  called  the 
remainder.  In  (b),  Art.  62,  11,352  is  the  dividend,  1892  is  the 
divisor.  6  is  the  quotient,  and  the  remainder  is  0.  Whenever  the 
remainder  is  0,  the  division  is  said  to  be  exact. 

64.  There  are  several  signs  for  division,  the  principal  one  being 
a  colon  (:)  or  a  colon  with  a  short  horizontal  line  between  the  dots 
(-5-).  When  either  of  these  two  signs  occurs  between  two 
numbers,  it  means  that  the  number  on  the  left  is  to  be  divided  by 
the  number  on  the  right:  thus.  24-4-6  means  that  24  is  to  be 
divided  by  6,  24  being  the  dividend  and  6  the  divisor.  In  some 
cases,  a  vertical  line  is  used  in  place  of  the  regular  sign  of  division; 
thus,  24j6.  The  vertical  line  is  seldom  used  between  two 
numbers :  it  is  most  frequenth"  used  when  the  product  of  several 

124  '   84  numbers  is  to  be  divided  by  the    product    of 

112  several  other  numbers;  thus,  the  product  of  124, 

49,  and  75  divided  by  the  product  of  84  and  1 12 
may  be  indicated  as  shown  in  the  margin,  the  product  of  the 
numbers  to  the  left  of  the  vertical  line  being  the  dividend,  and  the 
product  of  the  numbers  to  the  right  being  the  divisor.  Most 
commonly,  however,  in  cases  of  this  kind,  a  horizontal  line 
is  used,  the  number   or  numbers  above  the  fine  being  divided 

24 
by  the  number  or  numbers  below  it ;  thus,  -«■  is  read  24  over  6, 

i  q,    v   -i    i  i     r      i       124X49X75  ..    _.  ._ 

and  means  24  divided  by  G;  also,  — cj.  y  i  io —  means  that  the 


§1  MULTIPLICATION  AND  DIVISION  29 

product  of  124,  49,  and  75  is  to  be  divided  by  the  product  of  84 
and  112.  An  inclined  line  is  also  frequently  used;  thus  24/6 
means  24  divided  by  6. 

65.  Short  Division. — When  the  divisor  is  not  greater  than  12,  it 

is  customary  to  employ  what  is  called  short 

division.     The  process  is  best  illustrated 

8)5  4*3786372  by  an  example.     For  instance,  543,832:8  =  ? 

6  7  9  7  9  Write  the  divisor  to  the  left  of  the  dividend 

with  a  curved  line  between,  as  shown  in  the 

margin,  and  draw  a  straight  line  under  the 

dividend.     Since  8  is  greater  than  5,  the  left-hand  digit  of  the 

dividend,   consider  the  first  two  figures,   54.     Now  find  what 

number  multiplied  by  8  will  make  54  or  whose  product  subt  racted 

from  54  will  be  less  than  8 ;  since  8  X  6  =  48  and  8  X  7  =  56, 

this  number  is  6,  and  54  —  48  =  6.     Write  6  under  54  for  the 

first  figure  of  the  quotient,  and  also  write  6,  the  remainder,  above 

and  to  the  right  of  4,  as  shown.     This  last  6  belongs  to  the  same 

order  as  the  4  in  54  and  the  3  that  follows  4  is  of  the  next  lower 

order;  hence,  combine  the  6  and  3,  and  call  the  number  63.     Now 

find  what  number  multiplied  by  8  will  make  63  or  whose  product 

subtracted  from  63  will  be  less  than  8;  this  number  is  7,  since  8  X 

7  =  56  and  63  —  56  =  7.  Write  7  under  3  for  the  second  figure 
of  the  quotient,  and  also  write  7,  the  remainder,  as  shown.  The 
next  number  to  be  divided  is  78,  the  7  being  prefixed  to  the  fourth 
figure  of  the  dividend,  which  is  the  figure  of  the  next  lower  order. 
Here  8  X  9  =  72,  and  78  -  72  =  6;  write  9  under  8  for  the  third 
figure  of  the  quotient  and  write  6,  the  remainder,  as  shown. 
Prefixing  6  to  3,  the  next  figure  of  the  dividend,  8  X  7  =  56,  63 
—  56  =  7;  hence,  write  7  for  the  fourth  figure  of  the  quotient  and 
write  7,  the  remainder,  as  shown.  Finally,  8  X  9  =  72,  and 
there  is  no  remainder;  hence,  write  9  for  the  fifth  figure  of  the 
quotient.  As  there  are  no  more  figures  in  the  dividend,  the  quo- 
tient sought  is  67,979.  That  this  is  cornet  may  be  proved  by 
multiplying  the  quotient  by  the  divisor,  the  result  being  67979  X 

8  =  543832,  the  dividend. 

In  practice,  the  remainders  would  not  be  written  in  the  manner 
indicated  in  the  foregoing — they  would  be  simply  carried  men- 
tally. The  process  would  then  be  as  follows:  S  into  54,  6  times 
and  6  over;  8  into  63,  7  times  and  7  over;  8  into  78, 9 times  and 
6  over;  8  into  63,  7  times  and  7  over;  8  into  72,  9  times  and  no  re- 


30  ARITHMETIC  §1 

mainder.     Suppose  it  were  required  to  divide  the  above  number, 

543,832  by  9.     Say  9  into  54,  6  times  and 

—  nothing  over;  9  into  3  no  times;  9  into  38, 

4  times  and  2  over;  9  into  23,  2  times  and 

5  over;  9  into  52,  5  times  and  7  over.     Since  there  are  no  more 

figures  in  the  dividend,  the  quotient  is  60,425  and  7  remainder. 

That  by  result  is  correct  may  be  proved  by  multiplying  the 

quotient  by  the  divisor  and  adding  the  remainder  to  the  product; 

thus,  60425  X   9   =   543825,  and  543825  +  7  =  543832.     Now 

note  that  the  remainder  is  the  same  as  that  obtained  in  Art.  61  by 

adding  the  digits;  thus  5  +  4  +  3  +  8  +  3  +  2  =  25,  and  2  +  5 

=  7. 

For  reasons  that  will  be  explained  later,  it  is  customary  to  write 
the  remainder  over  the  divisor,  with  a  line  between,  and  annex 
this  expression  to  the  quotient.  In  the  last  example,  543832  -f-  9 
=  604251 .  Ans.  This  last  expression  may  be  read  sixty 
thousand  four-hundred-twenty-five  and  seven  over  nine. 


EXAMPLES 

(1)  197527  -j-  11  =  ?  Am.  17,957. 

(2)  527324  v7  =  ?  Ans.  75,332. 

(3)  900725  -f-  6  -  ?  Ans.   150,120*. 

(4)  ^P  =  ?  Ans.  4,125,263*. 

(5)  1580216/4  =  ?  Ans    395,054. 

(6)  4350688  +  3  =  ?  Ans.   1,450,2291. 

(7)  2072623/10  =  ?  Ans.  207,262T30. 

Note. —To  divide  a  number  by  10,  write  all  the  figures  except  the  last  for  the  quotient; 
the  last  figure  will  be  the  remainder  (see  example  7,  above). 

(8)  Since  there  are  12  inches  in  one  foot,  how  many  feet  are  equivalent  to 
237  inches?  Ans.  19^  feet. 

(9)  How  many  nickels  are  equal  in  value  to  $3.45,  one  nickel  being  equal 
to  5  cents?  Ans.  69  nickels. 

(10)  Twelve  of  anything  make  a  dozen,  and  twelve  dozen  make  a  gross. 
How  many  dozen  balls  of  twine  are  in  a  shipment  containing  5076  balls? 
also,  how  many  gross  were  in  the  shipment?  Ans.  423  dozen ;  35-jV  gross. 

(11)  There  are  8  pints  in  one  gallon;  how  many  gallons  are  equivalent  to 
22,222  pints?  Ans.  2777*  gallons. 

(12)  One  yard  is  equal  to  three  feet;  how  many  yards  are  contained  in 
63360  inches?  Ans.   1760  yards. 

(13)  A  quire  of  paper  contains  24  sheets.  How  many  quires  are  in  a  pile 
of  1784  sheets?  Ans.  74^  quires. 


§1  MULTIPLICATION  AND  DIVISION  31 

66.  Long  Division. — When  the  divisor  is  greater  than  12,  the 
process  called  long  division  is  used;  this  is  besl  explained  by  an 

exam  pie.     For  instance,  what  is 
64903358)386  the  quotient  when  64,903,358  is 

386  168143* ge     Ans-        divided  by  386  ?     Write  the  divi- 

2630  sor  to  the  right  of  the  dividend, 

2316  with  a  line   (either  straight  or 

3143  curved)  between,  and  draw  a  line 

30S8 

under    the    divisor,    as    shown. 

553 

ooc  Since  the  divisor  contains  3  fig- 
38b  .       ...  , 

— _  ures,  compare  it  with  the  number 

1544  made  up  of  the  first  3  figures  of 

~~1318  the  dividend,  in  this  case,  649, 

1158  which  call  the  first  trial  dividend. 

160  Note  that  649  is  larger  than  386, 

which  is  nearly  equal  to  400; 
calling  it  400,  it  is  seen  that  400  is  contained  in  649,  1  time,  and 
1  is  thus  the  first  figure  of  the  quotient,  which  is  written  under 
the  divisor.  Now  multiply  the  divisor  by  1,  the  figure  of  the 
quotient  just  found,  and  write  the  product  under  the  first  three 
figures  of  the  dividend,  draw  a  line  under  it,  and  subtract,  obtain- 
ing a  remainder  of  263.  Annex  to  this  remainder  the  next  figure 
of  the  dividend,  in  this  case  0,  and  divide  2630,  which  is  the  new, 
or  second,  trial  dividend,  by  386,  the  divisor,  which  call  400  as  be- 
fore, obtaining  6  for  the  second  figure  of  the  quotient.  Multiply 
the  divisor  by  6,  the  figure  of  the  quotient  last  found,  and 
write  the  product,  2316,  under  2630;  subtract  as  before,  obtain- 
ing 314  as  a  remainder,  to  which  annex  the  next  figure  of  the 
dividend,  3  in  this  case,  making  the  new,  or  third,  trial  dividend 
3143.  To  divide  3143  by  386,  note  that  8  X  400  =  3200,  a 
number  slightly  larger  than  3143;  but  as  386  is  smaller  than 
400,  try  8  for  the  third  figure  of  the  quotient.  Multiplying  386 
by  8,  the  product  is  3088,  which  subtracted  from  3143,  leaves  a 
remainder  of  55,  to  which  annex  the  next  figure  of  the  divi- 
dend, in  this  case  3,  making  the  new,  or  fourth,  trial  dividend 
553.  The  next  figure  of  the  quotient  is  evidently  1,  and  the  next, 
or  fifth,  trial  dividend  is  1675.  Dividing  1675  by  400,  the  n.  xt 
figure  of  the  quotient  is  4;  multiplying  386  by  1  and  subtracting 
the  product  from  1675,  the  remainder  is  131,  to  which  annex  the 
next  (in  this  case,  the  last)  figure  of  the  dividend,  obtaining  1318 
for  the  new,  or  sixth,  trial  dividend.     Dividing  1318  by  400,  the 


32  ARITHMETIC  |1 

next  figure  of  the  quotient  is  3;  the  remainder  after  multiplying 
the  divisor  by  3  is  160.  Aa  there  are  no  more  figures  in  the  divi- 
dend, the  division  of  160  by  386  is  indicated  by  writing  386  under 
160  with  a  line  between.     The  quotient,  therefore,  is  168,143£f  |?. 

In  the  foregoing,  the  number  400,  which  was  used  in  place  of 
386  to  determine  the  different  figures  of  the  quotient,  is  called  the 
trial  divisor.  If  the  second  figure  of  the  divisor  is  5  or  a  larger 
digit,  increase  the  first  figure  of  the  divisor  by  1,  use  the  result 
thus  obtained  as  a  trial  divisor,  and  proceed  as  in  short  division 
to  obtain  the  next  figure  of  the  quotient.  But.  if  the  second 
figure  of  the  divisor  is  less  than  5,  use  the  first  figure  of  the  divisor 
for  a  trial  divisor.  In  case  there  is  any  doubt  as  to  whether  the 
figure  of  the  quotient  so  obtained  is  correct,  multiply  the  second 
figure  of  the  divisor  by  the  figure  thus  obtained  in  the  quotient 
and  add  the  amount  to  be  carried  to  the  product  of  this  figure  and 
the  first  figure  of  the  divisor,  comparing  the  result  with  the  first 
two  figures  of  the  trial  dividend.  Thus,  in  the  foregoing  ex- 
ample, to  determine  whether  to  try  7  or  8  for  the  third  figure  of 
the  quotient  8  X  8  =  64,  8  X  3  =  24,  and  24  +  6  =  30;  since 
30  is  smaller  than  31,  the  first  two  figures  of  the  trial  dividend, 
3  for  the  third  figure  of  the  quotient.  It  may  be  remarked 
that  the  quotient,  16S143|f£,  is  read  one  hundred  sixty-eigh 
thousand  one  hundred  forty-three  and  one  hundred  sixty  over 
three  hundred  eighty-six. 

67.  Check  for  Division. — To  check  division,  cast  out  9's  from 
the  dividend,  divisor,  quotient,  and  remainder;  the  product  of 
the  remainders  for  the  divisor  and  quotient  plus  the  remainder 
for  the  remainder  should  equal  the  remainder  for  the  dividend, 
but  if  not.  a  mistake  has  been  made.  Thus,  in  the  example  of 
Art.  66,  disregarding  the  9's.  6  +  4  +  3 +  3  +  5  +  8  =  29.  and 
the  remainder  is  2  for  the  dividend.  The  remainder  for  the  divi- 
sor is  :;  +  8  +  6  =  17.  and   1  +7  =  8;  the  remainder  for  the 

quotient  L,l +  6  +  S  +  1 +4 +  3  = 

'30821  .     23,  aml2  +  3  =  o;t„.,(™ai„der 

:0s  for  the  remainder  is  1  +  6  =  / ;  then 

2lS0:;."i  8X5  =  40.  or  4,  and  4  +  7  =  11, 

1  +  1=2.  the  same  remainder  as 

348856  was  found  for  the  dividend;  hence, 

1851  86  the  work  is  probably  correct. 

J_i;  As     another     example,     divide 

107  1,529,918,746    by    43607.       Since 


§1  MULTIPLICATION  AND  DIVISION  33 

the  second  figure  of  the  divisor  is  3,  a  digit  smaller  than  .">,  use  4 
for  the  trial  divisor.  Since  the  first  5  figures  of  the  dividend 
make  a  smaller  number  than  the  five  figures  of  the  divisor,  use 
the  first  six  figures  of  the  dividend  for  the  first  trial  dividend. 
Since  4  is  contained  in  15,  3  times,  3  is  the  first  figure  of  the 
quotient.  The  second  figure  of  the  quotient  is  easily  seen  to 
be  5,  and  the  second  remainder  is  3673;  annexing  7,  the  next 
figure  of  the  dividend,  the  third  trial  dividend,  36737,  is  smaller 
than  the  divisor;  hence,  write  a  cipher  (0)  for  the  third  figure 
of  the  quotient,  and  annex  4,  the  next  figure  of  the  dividend, 
making  the  fourth  trial  dividend  367374.  While  36  -8-  4  =  9,  9 
is  evidently  too  large,  since  43  X  9  =  387;  consequently,  try  8 
for  the  fourth  figure  of  the  quotient.  The  fifth  figure  is  4,  and 
the  remainder  is  10758,  which  is  written  over  the  divisor,  as  shown. 
Applying  the  check,  1  +  5  +  2  +  1  +  8  +  7  +  4  +  6  =  34, 
and  3  +  4  =  7;  4  +  3  +  6  +  7  =  20,  or  2;  3  +  5  +  8  +  4  = 
20,  or  2;  1  +  7  +  5  +  8  =  21,  and  2+1=3;  then  2X2  =  4, 
and  4  +  3  =  7,  the  same  remainder  as  was  obtained  for  the  divi- 
dend; hence,  the  work  is  probably  correct. 

68.  Rule  I. — Write  the  divisor  to  the  right  of  the  dividend,  with 
a  line  between,  and  draw  a  line  under  the  divisor. 

II.  Determine  the  trial  divisor  as  previously  described  and  divide 
it  into  the  first  trial  dividend  for  the  first  figure  of the  quotient;  multi- 
ply the  divisor  by  this  figure,  subtract  the  product  from  the  trial  divi- 
dend, and  annex  the  next  figure  of the  dividend  for  a  new  trial  dividend. 
Divide  the  second  trial  dividend  by  the  trial  divisor  for  the  second 
figure  of  the  quotient.  Proceed  in  this  manner  until  all  the  figures 
of  the  dividend  have  been  used. 

III.  If  any  trial  dividend  is  smaller  than  the  divisor,  write  a 
cipher  for  the  corresponding  figure  of  the  quotient,  and  annex  the 
next  figure  of  the  dividend  for  a  new  trial  dividend. 

IV.  If  there  is  a  remainder,  write  it  over  the  divisor  with  a  line 
between,  and  annex  this  expression  to  the  quotient. 


EXAMPLES 

U)  Divide  31415927  by  4726.  I    B641  ^fl 

7v.il) 
(2)  40073S3G  -^  8018  =  ?  Ans.   -**»'*~^n, g 

w   519  519 


34  ARITHMETIC  §1 

CO 

(4)  Divide  43560  by  209.  Ans.  2082^. 

(5)  30159681  -f-  5307  =  ?  Ans.  5683. 
„.  4,396,652,679       _                                                           .        C1rtftrt33942 

^6)  ^6193         -  ?  AnS-  5100986193- 

(7)  Divide  2,189,404,900  by  29,950.  Ans.  73,102. 

(8)  According  to  Bessel,  the  diameter  of  the  earth  at  the  equator  is  41,- 

847,192  feet;  what  is  the  diameter  in  miles,  one  mile  containing  5280  feet? 

nnn r3192 
Ans.  7925^2^?»    miles. 

(9)  How  many  reams  of  480  sheets  each  are  contained  in  75,960  sheets  of 

120 
writing  paper?  Ans.  158jb~  reams- 

(10)  How  many  bales  of  rags  can  be  made  up  from  238,996  pounds  of 
rags,  if  the  bales  average  596  pounds  each?  Ans.  401  bales. 

(11)  12,656  pounds  of  paper  is  to  be  put  up  in  reams  of  25  pounds  each; 
how  many  reams  will  it  make?  Ans.  506/g  reams. 

(12)  The  freight  bill  on  a  shipment  of  pulpwood  called  for  payment  on 
246,782  pounds;  if  the  average  weight  of  a  cord  is  4450  pounds,  how  many 

cords  are  there?  Ans.  55.  .-„  cords. 

4450 


SOME  PROPERTIES  OF  NUMBERS 


DIVISIBILITY  OF  NUMBERS 

69.  As  previously  stated,  the  factors  of  the  product  of  two 
numbers  are  the  two  numbers  which,  when  multiplied  together 
produce  the  product.  If  more  than  two  numbers  are  multiplied, 
the  product  has  more  than  two  factors;  thus,  4X7X12X25  = 
8400,  and  8400  may  be  considered  to  have  as  its  factors  4,  7,  12, 
and  25.  Since  12  =  3  X  4,  and  25  =  5  X  5,  8400  also  has  as  its 
factors  3,  4,  4,  5,  5,  and  7,  because  3X4X4X5X5X7  = 
8400.  These  factors  may  be  combined  in  any  way  to  form  other 
factors;  thus,  3  X  5  =  15,  4  X  5  =  20,  and  4  X  15  X  20  X  7  = 
8400,  or  4  X  5  X  5  =  100,  and  3  X  4  X  7  X  100  =  8400,  etc. 

70.  A  multiple  of  a  number  (the  given  number)  is  a  certain 
number  of  times  the  given  number;  thus,  24  is  a  multiple  of  6, 
6  being  the  given  number,  because  4  times  6  is  24;  it  is  a  multiple 
of  8,  because  3  times  8  is  24;  it  is  a  multiple  of  12,  because  2  times 
12  is  24.  In  other  words,  any  number  that  can  be  expressed 
as  the  product  of  two  or  more  factors  is  a  multiple  of  any  one  of  the 


§1  SOME  PROPERTIES  OF  NUMBERS  35 

factors  or  of  the  product  of  two  or  more  of  its  factors.  For  in- 
stance, 7854  =  2  X  3  X  7  X  11  X  17;  it  is,  therefore,  a  multi- 
ple of  any  one  of  these  numbers  and  may  be  exactly  divided  by  any 
one  of  them,  and  when  it  has  been  divided  by  one  of  them,  the 
quotient  may  be  exactly  divided  by  any  one  of  the  remaining 
factors;  thus,  7854  ■*■  11  =  714,  and  714  may  be  exactly  divided 
by  any  of  the  remaining  factors,  since  714  -r  2  =  357;  357  -5-  3 
=  119;  119  -7-  7  =  17,  the  last  factor.  A  multiple  of  several  fac- 
tors may  be  exactly  divided  by  the  product  of  any  number  of 
those  factors;  thus,  7854  4-  11  X  17  =  7854  ■*■  187  =  42;  7854 
4-3X7X11  =  7854  4-  231  =  34;  etc. 

When  a  number  can  be  exactly  divided  by  another  number,  the 
first  number  is  said  to  be  divisible  by  the  second  number;  if, 
however,  there  is  a  remainder  after  the  division,  then  the  first 
number  is  not  divisible  by  the  second.  For  example,  84  is  divis- 
ible by  2,  3,  4,  6,  7,  12,  14,  21,  28,  and  42,  and  by  no  other  num- 
bers except  itself  (84)  and  1 ;  84  is  also,  of  course,  a  multiple  of 
these  numbers. 

71.  An  odd  number  is  one  whose  last  (right-hand)  figure  is  1,  3, 
5,  7,  or  9;  71,  423,  625,  1007,  1649  are  odd  numbers. 

An  even  number  is  one  whose  last  figure  is  0,  2,  4,  6,  or  8;  640, 
972,  1774,  31416,  and  2008  are  even  numbers. 

Any  even  number  is  divisible  by  2;  this  may  be  considered  as 
another  definition  of  an  even  number.  Thus,  any  of  the  above 
even  numbers  are  divisible  by  2. 

Any  number  ending  in  5  is  divisible  by  5;  thus,  635,  895,  etc. 
are  divisible  by  5. 

Any  number  ending  in  0  is  divisible  by  5  and  by  10;  thus,  640 
=  64  X  10  =  64  X  2  X  5.  Since  any  number  ending  in  0  is  a 
multiple  of  10,  it  has  for  two  of  its  factors  2  and  5;  it  is  therefore 
divisible  by  2,  5,  and  2  X  5  =  10.     See  Art.  70. 

72.  A  number  that  is  not  divisible  by  any  number  except  itself 
and  1  is  called  a  prime  number  or  a  prime;  all  prime  numbers 
except  2  are  odd  numbers,  since  any  even  number  is  divisible  by 
2.  The  prime  numbers  less  than  100  are  1,  2,  3,  5,  7,  11,  13,  17, 
19,  23,  29,  31,  37,  41,  43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97. 

A  number  that  is  divisible  by  some  number  other  than  itself 
and  1  is  called  a  composite  number.  Composite  numbers  may  be 
odd  or  even,  and  may  always  be  expressed  as  the  product  of  two 
or  more  prime  numbers;  thus,  84  =  2X2X3X7;  105  =  3X5 


36  ARITHMETIC  §1 

X  7;  69  =  3  X  23;  etc.  When  the  factors  are  prime  numbers, 
they  are  called  prime  factors. 

(a)  Any  number  ending  in  two  ciphers  is  divisible  by  4  and  by 
25;  thus,  62100  =  621  X  100  =  621  X  4  X  25,  since  100  =  4 
X  25. 

(6)  Any  number  ending  in  three  ciphers  is  divisible  by  8  and  by 
125;  thus,  621000  =  621  X  1000  =  621  X  8  X  125,  since  1000 
=  8  X  125. 

Since  100  =  2  X  50  =  20  X  5,  a  number  ending  in  two 
more  ciphers  is  divisible  by  2,  4,  5,  10,  20,  25,  and  50. 

Since  1000  =  2  X  500  =  4  X  250  =  5  X  200  =  8  X  125  = 
10  X  100  =  20  X  50  =  40  X  25,  a  number  ending  in  three  or  more 
ciphers  is  divisible  by  2,  4,  5,  8,  10,  20,  40,  50,  100,  125,  200,  250, 
and  500. 

(c)  If  the  sum  of  the  digits  of  any  number  is  a  multiple  of  3, 
the  number  is  divisible  by  3;  when  adding  the  digits,  it  is  not 
necessary  to  add  3,  6,  or  9.  Thus,  31416  is  divisible  by  3,  because 
1  +  4+1  =  6,  a  multiple  of  3 ;  similarly,  2350976  is  not  divisible 
by  3,  because  2  +  5  +  7  =  14,  which  is  not  a  multiple  of  3. 

(d)  If  the  number  is  even  and  the  sum  of  the  digits  is  a  multi- 
ple of  3,  the  number  is  divisible  by  6;  thus,  31416  is  divisible  by 
6,  because  it  is  even  and  the  sum  of  the  digits  is  a  multiple  of  3. 
Similarly,  790108  is  not  divisible  by  6,  because  7+1  +  8  =  1 6, 
which  is  not  a  multiple  of  3.  Had  the  number  been  780108, 
7  +  8  +  1  +  8  =  24,  a  multiple  of  3,  and  the  number  is  divisible 
by  6.     No  odd  number  is  divisible  by  6. 

(e)  If  the  sum  of  the  digits  is  a  multiple  of  9,  the  number  is 
divisible  by  9.  Thus,  2,072,322  is  divisible  by  9,  because  2+7 
+  2  +  3  +  2  +  2  =  18,  a  multiple  of  9;  and  3,263,016  is  not 
divisible  by  9,  because  3  +  2  +  6  + 3 +  1  +  6  =  21,  which  is 
not  a  multiple  of  9;  it  is  a  multiple  of  3,  however,  and  since  the 
number  is  even,  it  is  divisible  by  3  and  by  6. 

(/)  A  number  is  divisible  by  4  when  the  last  two  figures  are 
divisible  by  4;  thus,  4692,  543,836,  127,324,  etc.  are  all  divisible 
by  4,  because  92,  36,  and  24,  the  last  two  figures  of  each  number 
are  divisible  by  4.  But,  1742,  67,583,  98,782 ,  etc.  arc  not  divisible 
by  4. 

(g)  A  number  is  divisible  by  8  when  the  last  three  figures  are 
divisible  by  8;  1752,  57064,  31416,  etc.  are  divisible  by  8,  because 
752,  064,  and  416,  the  last  three  figures  of  each  number  are  divis- 
ible by  8.     But  1852,  57164,  and  31426  are  not  divisible  by  8. 


§1  SOME  PROPERTIES  OF  NUMBERS  37 

CANCELATION 

73.  When  one  composite  number  is  to  be  divided  by  another 

composite  number  or  when  the  product  of  several  numbers  is  to 

be  divided  by  the  product  of  several  other  numbers,  the  work 

can  frequently  be  shortened  by  employing  the  process  called 

cancelation.     Cancelation  means  dividing  out  or  canceling  equal 

factors  from  the  dividend  and  divisor.     For  example,  suppose  21G 

were  to  be  divided  by  36.     According  to  Art.  72,  216  is  divisible 

by  4  and  by  9,  and  36  is  also  divisible  by  4  and  by  9.     Indicating 

216 
the  division  by  -=«-,  divide  both  dividend  and  divisor  by  4,  and 

the  result  is  V;  again  dividing  both  dividend  and  divisor,  this 
time  by  9,  the  result  is  ?  =  6,  and  6  is  the  quotient  of  216  +  36. 

6 
54 

2,1b" 
In  practice,  the  work  would  be  performed  as  follows:   ...    =  6. 

1 
Here  216  is  divided  by  4,  a  line  is  drawn  through  it,  thus  striking 
out  or  canceling  216,  and  the  quotient,  54,  is  written  above  it; 
36  is  also  divided  by  4,  canceled,  and  9,  the  quotient,  is  written 
below  it.  Since  54  and  9  have  a  common  factor,  9,  54  is  divided 
by  9,  canceled,  and  the  quotient,  6,  is  written  above  it;  9  is  divided 
by  9,  canceled,  and  the  quotient,  1,  is  written  under  it.  Since  6 
■*-  1  =  6,  the  quotient  of  216  -5-  36  is  6.  The  same  result  might 
have  been  obtained  by  resolving  the  dividend  and  divisor  into 
their  prime  factors  and  canceling  those  that  are  common;  thus, 

216  _  ^XgX2X^X^X3  =  2X3  =  6.     When    the  quo- 

36  "  2  X  2  X  £  X  3 

tient  is  1,  it  is  not  customary  to  write  it  when  canceling. 

Canceling  the  same  factor  in  both  dividend  and  divisor  does  not 
alter  the  value  of  the  quotient. 

74.  If  there  are  several  factors  in  both  dividend  and  divisor,  the 
process  is  similar;  any  factor  common  to  both  dividend  and  divisor 
may  be  canceled.  For  example,  divide  L36  X  54  X  1 17  by  26  X  51 
X40.    Writing  the  dividend  over  the  divisor,  with  a  line  between, 

9 
17      '77        9 

w  x  h  x  w  =ox9  =8i  =  l6J 

^XolX40  5  5 

n     j$     5 


38  ARITHMETIC  §1 

By  Art.  72,  136  is  divisible  by  8;  40  is  also  a  multiple  of  8;  hence, 

cancel  136  and  40,  writing  the  quotients  as  shown.     Since  26 

and  54  are  divisible  by  2,  cancel  and  write  the  quotients  as  shown. 

By  Art.  72,  51  is  divisible  by  3,  and  since  27  is  a  multiple  of  3, 

cancel  and  write  the  quotients  as  shown.     Since  there  is  a  17 

above  the  line  and  another  below  it,  cancel  the   17's.     By  Art 

72,  117  is  divisible  by  9;  it  is  equal  to  9  X  13;  hence,  divide  the 

13  below  the  line  into  117.     As  there  are  no  more  common  factors, 

9X9 
the  original  expression  is  equal  to  — z — »  being  the  product  of  all 

o 

the  uncanceled  factors  above  the  line  divided  by  the  product  of 

all  the  uncanceled  factors  below  the  line. 

As  another  example,   what  is  the  value  of 

224  X  5700  X  189  X  85 
250  X  456  X  21  X  17  " 


Here, 


2 
If  9 

28        &U        68        17 


224  X  5700  X  1*9^x85  _ 

250  X  456  X  n  X  J$       ^  X  2  X  y      5U4' 

■2'>       •"■:       7 

By  Art.  72,  224  and  456  are  both  divisible  by  8;  hence,  cancel 
and  write  the  quotients  as  shown.  Cancel  the  10  in  250  and  5700, 
leaving  25  and  570.  Cancel  the  57  in  the  divisor  into  570  in  the 
dividend.  Cancel  5  in  25  and  10,  leaving  5  and  2,  and  cancel  the 
5  into  85,  leaving  17,  which  cancels  17  in  the  divisor.  189  and 
21  are  both  divisible  by  3,  and  63  is  a  multiple  of  7.  Canceling 
as  shown,  all  the  factors  in  the  divisor  have  been  canceled,  leaving 
28  X  2  X  9  =  504  for  the  quotient.  That  this  is  correct  may  be 
proved  by  actual  multiplication  and  division;  thus  224  X  5700  X 
189  X  85  =  20,511,792,000;  250  X  456  X  21  X  17  =  40,698,- 
000;  and  20,511,792,000  +  40,698,000  =  504. 

Rule. — Cancel  the  factors  that  are  common  to  both  dividend  and 
divisor,  and  divide  the  product  of  all  the  factors  that  remain  in  the 
dividend  by  the  product  of  all  the  factors  that  remain  in  the  divisor. 


§1  SOME  PROPERTIES  OF  NUMBERS  39 

EXAMPLES 

39  X  152  X  87  X  96      =  A       m 

y  '   19  X  6  X  29  X  42  X  13 


(2) 

(3) 


30X40X50X60X70  A        - 

15  X  25  X  35  X  45  X  55 

24  X  44  X  64  X  84  X  104  _  ^ns.  256. 

33  X  32  X  42  X  52 


231_X328J<_7200  =  Aris    773, 

^  '      21  X  64  X  525 


(5) 
(6) 


31416  X  55  X  192  =  ^  57Q 

121  X  56  X  85 
5236  X  630  X  128  X  192  =  ?  Am    99?, 

196  X  32  X  144  X  90 


ARITHMETIC 

(PART  1) 


EXAMINATION  QUESTIONS 

(1)  Express  in  figures  the  following  numbers:  (a)  ten  million 
nineteen  thousand  forty-two;  (6)  seventy  thousand  six  hundred 
five;  (c)  five  hundred  sixty-two  hundredths. 

(2)  Express  the  following  numbers  in  Roman  notation:  (a) 
4068;  (6)  44;  (c)  657,903;  (d)  1920;  (e)  1888. 

(3)  What  is  the  sum  of  $18.04,  $1.57,  $197.85,  $36.43,  and 
$360.52?  Ans.   $614.41. 

(4)  Seven  barrels,  with  their  contents,  weigh  respectively  297 
pounds,  417  pounds,  226  pounds,  388  pounds,  293  pounds,  185 
pounds,  and  313  pounds;  if  they  are  all  hoisted  in  an  elevator  at 
one  time,  and  the  elevator  weighs  2309  pounds,  together  with  two 
men,  one  weighing  148  pounds  and  the  other  187  pounds,  what  is 
the  total  weight  lifted?  Ans.  4763  pounds. 

(5)  From  a  pile  of  pulp  weighing  12,882  tons,  2057  tons  were 
removed  on  a  certain  day  and  3836  tons  on  another  day;  how 
many  tons  remained  in  the  pile?  Ans.  6989  tons. 

(6)  If  256,405  be  subtracted  from  a  certain  number  and  the 
remainder  is' 700,999;  (a)  what  is  the  number?  (6)  What  num- 
ber must  be  subtracted  from  1,001,010  to  give  a  remainder  of 

660,019?  Ans     I  (°)  95^04. 

A         \  (b)  339,991. 

(7)  What  is  the  product  of  461,  217  and  865.  Check  the  re- 
sult by  casting  out  9's.  Ana.  86,532,005. 

(8)  If  the  quotient  is  5077,  the  divisor  is  6345,  and  the  re- 
mainder is  2609,  what  is  the  dividend?  Check  by  casting  out 
9,s  Ans.  32,216,174. 

(9)  If  the  dividend  is  4,511,856,  the  quotient  is  2803,  and  the 
remainder  is  1829,  what  is  the  divisor?  Ans.   1609. 

(10)  There  are  5280  feet  in  a  mile;  (a)  how  many  miles  in 
408,806  feet?     (6)  How  many  feet  in  1894  miles? 

.         /(a)  77HH  miles. 
AnS'    \(6)   10,000,320  feet. 

41 


42  ARITHMETIC  §1 

(11)  There  are  five  numbers:  (1)  840,  (2)  231,  (3)  1728,  (4) 
2618,  and  (5)  1215;  (a)  which  of  these  numbers  are  odd?  (6) 
which  are  even?  (c)  which  are  divisible  by  3?  (d)  by  6;  (e)  by  9; 
(/)  by  12? 

(12)  Referring  to  the  last  question,  which  of  the  numbers  are 
divisible  (a)  by  2?  (6)  by  4?  (c)  by  8?  (d)  by  5? 

nQ,  ..    ,,,         ,         ,5236X9X45X26, 

(13)  Find  the  value  of  —  ^ —  by  cancelation. 

Ans.  51. 

(14)  A  hogshead  contains  63  gallons;  how  many  hogsheads  are 
equal  to  20,610  gallons?  Ans.  327TV  hogsheads. 

fin  tk    i+k        i        ,1309X1728X448, 

(15)  Find  the  value  of  v  88  X  154       y  cancelatl0n- 

Ans.  291T1. 


ARITHMETIC 

(PART  2) 


GREATEST  COMMON  DIVISOR  AND  LEAST 
COMMON  MULTIPLE 

76.  Greatest  Common  Divisor. — The  greatest  common  divisor 
of  two  or  more  numbers  is  the  largest  number  that  will  exactly 
divide  all  of  them.  Thus,  the  greatest  common  divisor  (usually 
abbreviated  G.C.D.)  of  36  and  48  is  12,  because  12  is  the  largest 
number  that  will  divide  36  and  48— it  is  the  largest  common 
factor  of  36  and  48;  also,  36  is  the  G.C.D.  of  72,  108,  and  144, 
because  it  is  the  largest  common  factor  of  those  numbers. 

To  find  the  G.C.D.  of  two  numbers,  find  the  product  of  their 
common  factors;  the  work  is  usually  arranged  as  follows:  Find  the 

G.C.D.  of  54  and  216.  Write  the 
M'  216  numbers  as  shown  in  the  margin, 


fh ?*  separating  them  by  a  comma,  and 


!>      4  then  proceed  as  in   short   division 

G.C.D.  =6  X  9  =  54  Ans.  Ry  ^    ^  ^  and  ^  ^  ^  d[_ 

visible  by  9;  divide  as  shown,  obtaining  6  and  24  for  the  quotients. 
Both  6  and  24  are  divisible  by  6,  the  quotients  being  1  and  4. 
The  only  factor  common  to  1  and  4  is  1,  which  has  no  effect  on 
the  G.C.D.  Since  54  and  216  have  the  common  factors  9  and  6, 
they  are  divisible  by  the  product  of  these  common  factors,  and 
the  G.C.D.  of  54  and  216  is  9  X  6  =  54.  To  prove  the  work, 
54  -=-  54  =  l,  and  216  -^  54  =  4. 

76.  To  find  the  G.C.D.  of  more  than  two  numbers,  proceed  in 
exactly  the  same  way  as  in  Art,  76,  remembering  that  the  divisors 
must  divide  all  the  numbers.     Thus,  find  the  G.C.D.  of  270, 

405,  675,  and  945.     Arrang- 

9  270,  405,  675,  945  ing  the  work  as  shown,  it  is 

5    30,  45,    75,  105  seen  that  all  the  numbers  are 

3  ~6J     9,    15,    2F  multiples  of  9;  hence,  divide 

~~2      3      1>,     7~  by  9.     The  quotients  are  all 

G.C.D.  =9X5X3  =  135.  Ans.  divisible  by  5;  hence,  divide 

43 


44 


ARITHMETIC 


§1 


391 

238 

1 

238 

153 

1 

153 

85 

1 

85 

68 

1 

68 

17 

4 

68 

0 

G.C. 

D.  = 

17 

Ans. 


by  5.  The  quotients  in  the  third  line  are  all  multiples  of  3; 
hence,  divide  by  3.  The  quotients  in  the  fourth  line  have  no 
common  factor;  therefore,  the  G.C.D.  is  9X5X3  =  135. 
To  prove  this,  divide  each  of  the  original  numbers  by  135;  the 
quotients  will  be  the  same  as  in  the  bottom  line. 

77.  It  frequently  happens  that  the  common  factors  are  not 
readily  seen;  in  such  cases,  proceed  as  in  the  following  example. 
Find  the  G.C.D.  of  238  and  391.     Write  the  larger  number,  and 

draw  a  vertical  line  on  its  right;  then 
write  the  smaller  number  and  draw 
another  vertical  line.  The  first  vertical 
line  is  the  sign  of  division,  but  the 
second  vertical  line  is  merely  a  line  of 
separation.  Now  divide  391  by  238, 
and  write  the  quotient  to  the  right  of 
the  second  vertical  line.  The  remainder 
is  153,  which  is  divided  into  238,  the  quotient  being  written  under 
the  first  quotient.  The  remainder  from  the  second  division  is 
85,  which  is  divided  into  153,  giving  a  remainder  of  68,  the  quo- 
tient being  written  under  the  preceding  quotient.  Dividing  85 
by  68,  the  remainder  is  17.  Lastly,  68  divided  by  17  gives  a 
quotient  of  4  and  a  remainder  0;  and  17,  the  divisor  that  gave  the 
remainder  0,  is  the  G.C.D.  of  238  and  391. 

As  another  example,  find  the  G.C.D.  of  31,416  and  1,400,256. 
Arranging  the  work  as  before,  the  first 
quotient  is  44,  and  the  remainder  is 
17.952.  which  is  divided  into  31,416. 
The  second  quotient  is  1,  and  the  second 
remainder  is  13,464,  which  is  divided 
into  17,952.  The  third  quotient  is  1, 
and  the  third  remainder  is  4,488,  which 
is  contained  in  13,464  3  times,  and  there 
is  no  remainder.  Therefore,  the  G.C.D.  of  31,416  and  1,400,256 
is  4488. 

It  is  seldom  necessary  to  find  the  G.C.D.  of  more  than  two 
numbers.  The  first  method  should  be  used  whenever  possible; 
but,  when  the  numbers  are  large  or  the  common  factors  are  not 
apparent,  then  use  the  second  method.  Observe  that  the  process 
must  be  carried  out  until  a  remainder  of  0  is  obtained.  If  the 
numbers  have  no  common  divisor,  the  G.C.D.  will  be  1.  In 
such  case,  the  numbers  are  said  to  be  prime  to  each  other,  though 


1400256 

125664 

143616 

125664 

17952 

13464 

4488 

31416 
17952 
13464 

13464 

o 

44 
1 
1 
3 


Ans. 


§1  LEAST  COMMON  MULTIPLE  45 

both  may  be  composite  numbers.     For  example,  91   and  136  are 

both  composite,  but  they  are  prime  to 
each  other,  as  is  evident  when  the 
greatest  common  divisor  is  found  to  be 
1,  in  accordance  with  the  work  as  shown 
in  the  margin. 


136 

91 

1 

91 

90 

2 

45 

1 

45 

45 

0 

G.C.D.  =  1. 

Ans. 

EXAMPLES 

(1)  Find  the  G.C.D.  of  40,  00,  SO,  and  100.  Ans.  20. 

(2)  Find  the  G.C.D.  of  70,  175,  210,  and  245.  Ans.  35. 

(3)  Find  the  G.C.D.  of  2387  and  1519.  Ana.  217. 

(4)  Find  the  G.C.D.  of  4059  and  6390.  Ans.  9. 

(5)  Find  the  G.C.D.  of  11433  and  444.  Ans.  111. 

(6)  Find  the  G.C.D.  of  364089  and  457368.  Ans.  3009. 

(7)  Find  the  G.C.D.  of  1016752  and  991408.  Ans.  176. 


78.  Least  Common  Multiple. — A  common  multiple  of  several 
numbers  is  a  number  that  is  divisible  by  those  numbers;  thus, 
840  is  a  common  multiple  of  40  and  56. 

The  least  common  multiple  of  two  or  more  numbers  is  the 
smallest  number  that  is  divisible  by  those  numbers;  thus,  280  is 
the  least  common  multiple  of  40  and  56;  it  is  the  smallest  number 
that  is  divisible  by  both  40  and  56. 

The  product  of  two  or  more  numbers  is  a  common  multiple  of 
the  numbers;  for  the  want  of  a  special  name,  this  will  be  called 
the  prime  multiple  of  the  numbers.  The  prime  multiple  may  be 
the  least  common  multiple,  and  will  be  such  if  no  two  of  the 
numbers  have  a  common  factor;  thus,  the  prime  multiple  of  91 
and  36  is  91  X  36  =  3276,  and  this  is  also  the  least  common 
multiple  of  91  and  36.  The  words  least  common  multiple  are 
usually  abbreviated  to  L.C.M.  If  the  two  numbers  have  a  com- 
mon factor,  it  may  be  canceled  from  the  numbeiB,  and  the  L.C.M. 
will  then  be  the  product  of  this  common  factor  and  the  remaining 
factors  of  the  numbers.  Thus,  56  and  72  have  t  he  common  factor 
8;  dividing  56  and  72  by  8,  the  quotients  are  7  and  9,  reaped  Ively ; 
then  the  L.C.M.  is  8X7  X  9  =  56X  9  =  7  X8  X9  =  7  X 
72  =  504.  Note  that  504  is  equal  to  56  X  9  and  to  72  X  7; 
hence,  it  is  divisible  by  both  56  and  72.  Since  8,  the  common  fac- 
tor, is  the  greatest  common  factor,  that  is,  the  G.C.D.,  of  56  and 
72,  it  follows  that  the  L.C.M.  of  two  numbers  may  be  found  by 


46 


ARITHMETIC 


§1 


dividing  one  of  the  numbers  by  their  G.C.D.  and  multiplying  the 
other  number  by  the  quotient.  For  instance,  referring  to  ex- 
ample (3)  at  the  end  of  Art.  77,  the  G.C.D.  of  2387  and  1519  is 
217;  1519  +  217  =  7;  then  2387  X  7  =  16,709,  the  L.C.M.  of 
1519  and  2387.  Or,  2387  -f-  217  =  11,  and  1519  X  11  =  16,709. 
79.  To  find  the  L.C.M.  of  more  than  two  numbers,  proceed  in 
much  the  same  way  as  in  finding  the  G.C.D.  An  example  will 
illustrate  the  process.  Find  the  L.C.M.  of  16,  24,  30,  and  32. 
Arrange  the  work  as  shown.  Since  all  the  numbers  contain  the 
common  factor  2,  divide  it  out.     In  the  first  line  of  quotients, 


2 

16, 

24, 

30, 

32 

4 

8, 

12, 

15, 

16 

2 

2, 

3, 

15, 

4 

3 

1, 

3, 

15, 

2 

1, 

1, 

5, 

2 

L.C.M.  =2X4X2X3X5X2=  480.  Ans. 
three  of  the  numbers  are  multiples  of  4(=  2  X  2),  but  the  other 
number  contains  no  factor  in  common  with  4;  hence,  divide  by  4. 
The  second  line  of  quotients  contains  two  numbers  that  are 
multiples  of  2;  hence,  divide  by  2.  The  third  line  of  quotients 
contains  two  numbers  that  are  multiples  of  3;  hence,  divide  by  3. 
The  numbers  in  the  fourth  line  of  quotients  are  all  prime  to  each 
other,  and  no  further  division  is  possible.  The  product  of  the 
divisors  and  the  factors  in  the  last  line  of  quotients  is  the  L.C.M., 
which  is  equal  to  2X4X2X3X5X2  =  480. 

When  finding  the  L.C.M.,  divide  out  all  factors  that  are  com- 
mon to  two  or  more  numbers  until  a  row  of  quotients  is  obtained 
that  are  prime  to  one  another.  As  another  example,  find  the 
L.C.M.  of  15,  27,  55,  and  99.     There  is  no  factor  common  to  all 


5 

15,  27,  55,  99 

11 

3,  27,  11,  99 

3 

3,  27,    1,    9 

3 

1,    9,     1,    3 

1,    3,     1,     1 

L.C.M.  =5X11X3X3X3  =  1485.  Ans. 
the  numbers;  but,  since  15  and  55  are  multiples  of  5,  divide  by  5. 
Since  11  and  99  are  multiples  of  11,  divide  by  11.  The  remainder 
of  the  work  is  evident.  Note  that  whenever  a  number  is  not 
divisible  by  a  divisor,  the  number  is  brought  down  into  the  line 
of  quotients.     The  L.C.M.  is  5  X  11  X  3  X  3  X  3'  =  1485. 


§1 


LEAST  COMMON  MILTHM  1. 


47 


If  the  numbers  are  such  that  their  factors  are  not  apparent,  find 
the  L.C.M.  of  two  of  them,  and  then  find  the  L.C.M.  of  this  result 
and  the  third  number;  then  the  L.C.M.  of  the  second  result  and 
the  fourth  number,  and  so  on;  the  last  result  will  be  the  L.C.M.  of 
all  the  numbers.  For  instance,  to  find  the  L.C.M.  of  893,  1387, 
and  1121,  first  find  the  L.C.M.  of  893  and  1387  (or  1121).  Pro- 
1  ceeding  as  described  in  Art.  78,  first  find  the 
l  G.C.D.  of  893  and  1387.  The  work  is 
1         shown  in  the  margin,  and  the  G.C.D.  is  19. 

4  Then,   893  +  19  =  47,   and    1387  X  47  = 

5  65189,  the  L.C.M.  of  893  and  1387.  Now 
find  in  the  same  way  the  L.C.M.  of  1121 
and   65,189.     The   work    for   this   is   also 

shown  in  the  margin.     The  G.C.D.  of  these  two  numbers  is  19; 

1121  -7-  19  =  59;    and    65189  X  59  =  3,846,141,    which    is    the 

L.C.M.  of  893,  1387,  and  1121. 

It  may  here  be  remarked  that 
the  greatest  common  divisor  and 
the  least  common  multiple  are  of 
importance  in  connection  with 
the  reduction,  addition,  and 
subtraction  of  fractions,  as  will 
shortly  appear. 


1387 

893 

893 

494 

494 

399 

399 

380 

95 

19 

95 

0 

65189 

5605 

~9139 

8968 

171 

95 

76 

76 

0 

1121  -=-  19  =  59 
65189  X  59  =  3846151 


1121 

1026 

95 

76 

19 


58 
6 
1 
1 
4 


Ans. 


(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 


Find  the  L.C.M. 
Find  the  L.C.M. 
Find  the  L.C.M. 
Find  the  L.C.M. 
Find  the  L.C.M. 
Find  the  L.C.M. 
Find  the  L.C.M. 


EXAMPLES 

of  28,  49,  63,  and  84. 
of  12,  14,  16,  and  18. 
of  15,  20,  25,  and  30. 
of  4,  8,  12,  16,  and  20. 
of  1955  and  4403. 
of  119,  204,  272. 
of  442,  234,  and  1001. 


Ans.   1764. 

Ans.   1008. 

Ans.   300. 

Ana.   240. 

Ana.   506,346. 

Aim.  5712. 

An*.   306,306. 


48  ARITHMETIC  §1 

FRACTIONS 


DEFINITIONS 

80.  When  an  integer  or  a  unit  is  divided  into  equal  parts,  one  or 
more  of  these  parts  is  called  a  fraction  of  the  integer  or  unit.  If, 
for  example,  a  straight  stick  be  cut  into  two  pieces  of  the  same 
length,  one  piece  is  equal  to  the  other,  and  either  is  called  one- 
half  of  the  stick.  If  the  stick  is  cut  into  three  equal  pieces  or 
parts,  one  of  them  is  called  one-third  of  the  stick;  if  cut  into  four 
equal  parts,  one  of  them  is  called  one-fourth  of  the  stick;  if  cut  into 
five  equal  parts,  one  of  them  is  called  one-fifth  of  the  stick;  etc. 
The  expressions  one-half,  one-third,  one-fourth,  etc.  are  fractions. 
More  than  one  part  is  denoted  by  writing  the  number  before  the 
name  of  the  part;  thus,  two-thirds  means  two  one-thirds,  three- 
fourths  means  three  one-fourths,  etc. 

81.  To  express  a  fraction  with  figures,  it  is  necessary  to  write 
two  numbers,  one  to  show  into  how  many  parts  the  integer  or 
unit  has  been  divided,  and  the  other  to  show  how  many  of  these 
parts  are  taken  or  considered.     For  instance, 

\  means  one-half,  and  indicates  one  of  two  equal  parts 
\  means  one-third,  and  indicates  one  of  three  equal  parts 
\  means  one-fourth,  and  indicates  one  of  four  equal  parts 
f  means  two-thirds,  and  indicates  two  of  three  equal  parts 
\  means  four-fifths,  and  indicates  four  of  five  equal  parts 
yV  means  seven-twelfths,  and  indicates  seven  of  twelve  equal 
parts,  etc.,  etc. 

82.  It  will  be  observed  that  a  fraction  is  expressed  in  one  of  the 
ways  used  to  indicate  division.  The  number  below  the  line  is 
called  the  denominator,  because  it  denominates,  or  names,  the 
number  of  parts  into  which  the  integer  has  been  divided. 

The  number  above  the  line  is  called  the  numerator,  because  it 
numerates,  or  counts,  the  number  of  the  equal  parts  that  are  taken 
or  considered. 

83.  An  expression  like  f  may  be  interpreted  in  two  ways: 
First,  it  is  3  times  \  of  a  unit.  For  example,  one  dollar  is  equal 
to  100  cents;  one-fourth  of  a  dollar  is  100  4-  4  =  25  cents;  and 
three-fourths  of  a  dollar  is  3  X  25  cents  =  75  cents.  Here  one 
dollar  is  the  unit.  Second,  it  is  one-fourth  of  three  times  the  unit. 
If  the  unit  is  one  dollar,  three  times  the  unit  is  3  dollars  or  300 
cents,  and  one-fourth  of  3  times  the  unit  is  300  cents  -5-  4  =  75 


§1  FRACTIONS  49 

cents,  the  same  result  as  before.     In  the  first  case,  f  is  a  fraction; 
in  the  second  case,  it  is  an  indication  of  division. 

84.  Except  when  the  denominator  is  1,  2  or  3,  a  fraction  is 
read  by  pronouncing  the  name  of  the  numerator  and  then  pro- 
nouncing the  name  of  the  denominator  after  adding  ths;  thus,  f 
is  read  three-sevenths,  ]l  is  read  eleven-sixteenths,  j2\  is  read 
forty-nine  one-hundred-twenty-fifths,  etc.  But  I  \  is  read  t  hirteen- 
twenty-firsts ;  f  I  is  read  twenty-nine-f orty-scconds ;  1%  is  read 
thirty-seven  fifty-thirds,  etc. 

85.  If  two  fractions  have  the  same  numerator,  but  a  different 
denominator,  the  fraction  whose  denominator  is  the  smaller  is 
the  larger;  thus  f  is  larger  than  f,  because  one-fifth  of  anything 
is  smaller  than  one-fourth  of  it;  hence,  3  one-fifths  is  smaller  than 
3  one-fourths  For  example,  three-fourths  of  a  dollar  is  75  cents, 
but  three-fifths  of  a  dollar  is  60  cents,  since  one-fifth  of  a  dollar 
is  100  cents  4-  5  =  20  cents,  and  3  X  20  cents  =  60  cents. 

86.  The  numerator  of  a  fraction  may  be  greater  than  the  de- 
nominator, in  which  case,  the  value  of  the  fraction  is  found  by 
dividing  the  numerator  by  the  denominator;  thus,  V  =  4,  the 
value  of  the  fraction;  ~<r  =  2,  the  value  of  the  fraction;  etc.  If 
the  numerator  is  equal  to  the  denominator,  the  value  of  the  fract  i<  m 
is  1;  thus,  it  =  1,  |-  =  1,  etc.  This  is  evident,  since  if  a  dollar 
is  divided  into,  say,  5  equal  parts,  5  of  these  parts  make  up  the 
dollar.  If  the  numerator  is  less  than  the  denominator,  the 
value  of  the  fraction  is  less  than  1. 

87.  A  proper  fraction  is  one  whose  numerator  is  smaller  than 
its  denominator;  its  value  is  always  less  than  1.     Thus,  ^'. 

i  c,  etc.  are  proper  fractions. 

An  improper  fraction  is  one  whose  numerator  is  equal  to  or 
greater  than  its  denominator.  Thus,  |,  "\,  -.',',  etc.  are  impro- 
per fractions. 

When  it  is  not  desired  to  specify  the  numerator  and  denomina- 
tor separately,  they  are  called  the  terms  of  the  fraction;  thus,  the 
terms  of  the  fraction  f  J  are  23  and  42. 

When  a  fraction  is  joined  to  an  integer,  as  in  the  expression 
14f,  the  expression  is  called  a  mixed  number.  Here  14|  means 
14  and  f  more;  it  has  the  same  meaning  as  14  -f-  §.  In  reading 
a  mixed  number,  the  word  and  is  used  as  above,  but  the  plus  Bign 
is  always  understood,  though  not  written  or  spoken;  thus,  5£  is 
read  five  and  two-thirds,  and    means  5  +   }•     -Mixed  numbers 


50  ARITHMETIC  §1 

always  occur  whenever  the  dividend  is  not  a  multiple  of  the  divi- 
sor; for  instance,  the  quotients  found  in  examples  (3),  (4),  (6),  and 
(7)  of  Art.  65  are  mixed  numbers. 

88.  In  printing  and  writing,  in  order  to  save  space,  fractions  are 
frequently  expressed  by  using  the  inclined  line  instead  of  the 
horizontal  line;  thus,  ?i  =  £,  113/147  =  Hf,  etc.  In  such 
cases,  a  hyphen  is  sometimes  written  between  the  integer  and  the 
fraction  of  a  mixed  number;  thus,  14-3/8  =  14%;  the  hyphen 
shows  that  the  fraction  belongs  to  the  integer. 

When  it  is  desired  to  indicate  that  the  fraction  is  to  be  pro- 
nounced and  it  is  not  desired  to  write  the  name  in  full,  the  fraction 
is  written  as  usual  and  the  ending  of  the  name  of  the  denominator 
is  annexed;  thus,  %rds,  K2^hs,  1^2<is,  2%ists,  mean  two-thirds, 
seven-twelfths,  fifteen  twenty-seconds,  twenty-three  thirty-firsts, 
etc.  The  onhr  exception  is  \,  which  is  always  read  one-half,  and 
is  always  so  written  and  printed. 


REDUCTION  OF  FRACTIONS 

89.  To  reduce  a  fraction  is  to  change  its  form  without  changing 
its  value;  to  change  its  form  means  to  alter  its  numerator  and  de- 
nominator. It  was  shown  in  Arts.  82  and  83  that  a  fraction  may 
be  regarded  as  an  expression  of  division,  the  numerator  being  the 
dividend  and  the  denominator  the  divisor.  In  Art.  73,  it  was 
shown  that  canceling  the  same  factor  in  both  dividend  and  divisor 
does  not  alter  the  value  of  the  quotient;  hence,  dividing  both 
numerator  and  denominator  of  a  fraction  by  the  same  number 
does  not  alter  the  value  of  the  fraction.  For  example,  yf  = 
y\  =  h  Here  both  18  and  24  are  first  divided  by  2,  the  quo- 
tients being  9  and  12,  respectively.  Since  9  and  12  contain  the 
common  factor  3,  divide  both  9  and  12  by  3,  the  quotients  being 
3  and  4,  respectively.  The  numerator  and  denominator  of  the 
given  fraction  might  both  have  been  divided  by  their  greatest 
common  divisor,  6,  and  the  same  final  result,  f ,  obtained. 

That  2~fths  =  fths  is  easily  shown.  Thus,  a  gross  is  144; 
Htha  of  a  gross  is  18  times  ^th  of  144:  s^th  of  144  is  144  +  24 
=  6,  and  4-|ths  is  18  X  6  =  108.  But  fths  of  a  gross  is  3  times 
ith  of  144;  1th  of  144  =  36,  and  fths  of  144  is  3  X  36  =  108.  In 
the  same  way,  it  is  shown  that  iVths  0I"  144  is  108.     Therefore, 

1  8     _       9       _     3 

ST  —  TT  —  f  • 


§1  FRACTIONS  51 

90.  Multiplying  both  numerator  and  denominator  by  the  same 

number  does  not  alter  (change)  the  value  of  the  fraction;  this  is 

evident,  since  both  terms  of  the  new  fraction  may  be  divided  by 

the  number  used  as  a  multiplier,  thus  obtaining  the  original  frac- 

3X6 
tion.     For  instance,    ^tv^  =  24,  which,  as  has  just  been  shown 

2X7 
is  equal  to  f .     Similarly,  _  -     _  =  3-5-. 

o  X  1 

91.  When  the  fraction  is  reduced  to  lower  terms,  that  is,  when 
the  numerator  and  denominator  are  made  smaller  by  division, 
the  process  is  called  reduction  descending.  When  the  fraction  is 
reduced  to  higher  terms,  that  is,  when  the  numerator  and  de- 
nominator are  made  larger  by  multiplication,  the  process  is  called 
reduction  ascending. 

92.  When  the  fraction  has  been  reduced  by  division  until  the 
numerator  and  denominator  are  prime  to  each  other  (have  no 
common  factor),  the  fraction  is  said  to  be  in  its  lowest  terms; 
thus,  f,  tV,  2T>  etc.  are  fractions  in  their  lowest  terms.  When 
a  fraction  in  its  lowest  terms,  it  is  in  its  simplest  form. 

93.  To  reduce  a  fraction  to  its  lowest  terms,  cancel  all  factors 
that  are  common  to  the  numerator  and  denominator.  If  the 
factors  are  not  apparent,  and  it  is  desired  to  make  sure  that  the 
fraction  is  in  its  lowest  terms,  find  the  greatest  common  divisor, 
if  any,  of  the  numerator  and  denominator  and  divide  both  terms 
by  it. 

Example  1. — Reduce  iff  to  its  lowest  terms. 

Solution. — Since  both  terms  are  multiples  of  4,  divide  them  by  4,  and 
^|!=  ||.  Both  terms  of  the  new  fraction  are  multiples  of  7;  hence,  divid- 
ing by  7,  If  =  §.  Both  terms  of  the  new  fraction  arc  multiples  of  3; 
hence,  dividing  by  3,  |  =  §.  Since  the  terms  are  now  prime  to  each 
other,  the  fraction  is  in  its  lowest  terms.  In  practice,  the  work  would  be 
arranged  as  follows:  -^f-f  =  ||  =  f  =  |.  Ana. 

Example  2. — Reduce  lllll  to  its  simplest  form. 

Solution. — Since  both  terms  are  multiples  of  8  (see  Art.  72  >,  divide  them 
by  8,  and  lif^f  =  §f  jf.  Apparently,  the  terms  have  no  common  factor, 
but  to  make  certain,  apply  the  process  of  finding  the  G.C.D.  of  0409  and 
8671;  in  this  case,  the  G.C.D.  is  377.  Dividing  both  terms  by  377,  |g$f 
=  ?%.  Ans. 

94.  To  reduce  a  fraction  to  another  fraction  having  a  given 
denominator,  divide  the  given  denominator  by  the  denominator 
of  the  given  fraction  and  multiply  the  terms  of  the  fraction  by 
Che  quotient.     For  example,  to  reduce  f  to  a  fraction  having  96 


52  ARITHMETIC  §1 

.3X12       36       , 
for    its   denominator;    96  -5-  8  =  12,    and  q  y  io  ~  Q(r     ^ns- 

The  reason  for  dividing  the  given  denominator  by  the  denomina- 
tor of  the  given  fraction  is  evident,  since  the  quotient  must  be  the 
number  by  which  the  denominator  of  the  given  fraction  must  be 
multiplied  in  order  to  equal  the  given  denominator.  This  opera- 
tion is  of  importance  in  adding  and  subtracting  fractions. 

95.  To  reduce  an  integer  to  an  improper  fraction  having  a  given 
denominator,  multiply  the  integer  by  the  given  denominator,  and 
the  product  will  be  the  numerator  of  a  fraction  having  the  given 
denominator.  For  instance,  reduce  7  to  a  fraction  having  12  for 
its  denominator.  Here  7  X  12  =  84,  and  the  required  fraction 
is  fi-  That  this  result  is  correct  may  be  proved  by  dividing  the 
numerator  by  the  denominator,  the  quotient  being  7,  the  original 
number.     This  result  may  also  be  obtained  in  another  way;  thus, 

7  is  evidently  equal  to  t,  and  y        .  ~  =  y«j. 

96.  To  reduce  a  mixed  number  to  an  improper  fraction,  multi- 
ply the  integral  part  by  the  denominator  of  the  fraction,  add  the 
numerator  to  the  product,  and  write  the  sum  over  the  denomina- 
tor. This  is  evidently  correct,  since  the  mixed  number  is  obtained 
by  dividing  the  dividend  (numerator)  by  the  divisor  (denomina- 
tor), and  the  remainder  is  written  over  the  divisor.  Thus,  reduce 
14f  to  an  improper  fraction.  Here  14  X  8  =  112;  112  +  3 
=  115;  hence,  14f  =  3J^A.  Regarding  x^-  ths  as  an  indication  of 
dividing  115  by  8,  115  +  8  =  14f. 

97.  A  common  denominator  of  two  or  more  fractions  is  a  com- 
mon multiple  of  the  denominators  of  the  fractions;  and  the  least 
common  denominator  is  the  least  common  multiple  of  the  denomi- 
nators. For  instance,  the  least  common  denominator  of  f ,  \, 
and  f  is  24,  because  24  is  the  L.C.M.  of  3,  4,  and  8. 

98.  To  reduce  two  or  more  fractions  to  fractions  having  a  least 
common  denominator,  find  the  L.C.M.  of  the  denominators;  then, 
by  the  method  of  Art.  94,  reduce  each  fraction  to  a  fraction  hav- 
ing this  denominator. 

Example  1. — Reduce  f,  \,  and  f  to  fractions  having  a  least  common 
denominator. 

Solution.— The  L.C.M.  of  3,  4,  and  8  is  24;  |  =  1%;  \  =  &;  and  f 
=  l'].  Am. 

Example  2. — Reduce  U,  §•$,  and  Jf-j  to  fractions  having  a  least  common 
denominator. 


§1  FRACTIONS  53 

Solution.— The  L.C.M.  of  69  and  92  is  276;  the  L.C.M.  of  276  and  161  is 
1932,  which  is  therefore  the  L.C.M.  of  69,  92,  and  161.  Then,  1932  +  69 
=  28,  and  ft  X -H  =  ifM;  '-,:;-  -92  =  21>  and  H  x  H  =  ftf$;  1932 

-=-  161  =  12,  and  $-$§  X  J^  =  i'ltl'     Therefore,  the  required  fractions  are 
HM.  HH.  and  Htf  An*. 

Example  3. — T^Tiich  fraction  is  the  larger,  -,"&  or  {$$1 

Solution. — To  determine  which  is  the  larger,  reduce  them  to  a  common 
denominator;  then  the  one  that  has  the  greater  numerator  is  the  larger. 
Since  {1%  =  la!  reduce  ^83\  and  U  to  a  common  denominator,  preferably, 
the  least  common  denominator.  Since  133  and  39  have  no  common  factor, 
their  L.C.M.  is  their  product,  or  133  X  39  =  5187;  then  VsV  =  ffjf,  and 
||  =  |f  |f.     Therefore,  {% {?-  is  the  larger  fraction.  Ans. 

Example  4. — There  are  1760  yards  in  a  mile;  what  fraction  of  a  mile  is 
550  yards? 

Solution. — Since  there  are  1760  yards  in  one  mile,  the  number  of  miles 
or  parts  of  a  mile  in  550  yards  is  550  ■*■  1760  =  VWs  =  -j'^r  =  fV« 
Therefore,  550  yards  is  i^ths  of  a  mile.     Ans. 

Example  5.— Which  is  the  greater  500  yards  or  ^ftths  of  a  mile? 

Solution. — Since  there  are  1760  yards  in  a  mile,  500  yards  =  vtW 
=  T5T°F  =  2%8  mile;  -jW  -  H  mile-  Reducing  these  fractions  to  a  com- 
mon' denominator,  the  L.C.M.  of  88  and  53  is  88  X  53  =  4664;  |{ 
=  i||0(  and  ||  =  132  5  Therefore,  500  yards  is  a  little  greater  than 
-jV^ths  of  a  mile.  Ans. 


EXAMPLES 

(1)  Reduce  to  its  lowest  terms  Hf.  Ans.     -,t. 

(2)  Reduce  to  its  lowest  terms  f  j  1 .  Ans.  \  j . 

(3)  Reduce  to  its  lowest  terms  j :,;" .  Ans.     £&. 

(4)  Express  11  as  a  fraction  having  24  for  its  denominator.         Ans.   -,V- 

(5)  Reduce  12.^6  to  an  improper  fraction.  Ans.   V,.."- 

(6)  Reduce  SS2\U  to  an  improper  fraction.  Ans.   '  ",\V  '  • 

(7)  Which  is  the  larger r9y°s  or|2 3?  Ans-  Ht« 

(8)  Which  is  the  larger  U  or  |f?  Ans.  Jf. 

(9)  Which  is  the  larger  ||  or  Ans.  #. 

(10)  Reduce  to  their  least  common  denominator  |,  {j,  ,',,  and  ," .. 

Ans.   ',;.  | 

(11)  Reduce  to  their  least  common  denominator  j, ',,  ';,  ri.and  [;;. 

Ans. 

(12)  A  ream  of  writing  paper  contains    ISO  sheets;  328  sheets  is  what 
fraction  of  a  ream?  Ans.  J'th  ream. 

(.13)  How  many  sheets  arc  equal  to  ftths  of  a    nam  of  writing  paper? 

Ans.  280  sheets. 

(14)  Which  is  the  larger,  200  sheets  of  writing  paper  or  |}th  ream? 

.1  as.   |}th  ream. 

(15)  How  many  sheets  difference  are  there  between  400  sheets  of  writing 
paper  and  fjda  ream?  Ans.  5  sheets. 

(16)  If  a  car  of  sulphur  weighs  40,000  pounds,  what  fraction  of  B  ear-load 
is  26,800  pounds?  Ans.    ;;„:0th  car-load. 


54  ARITHMETIC  §1 

ADDITION  OF  FRACTIONS 

99.  The  sum  of  -/V  and  A  is  }£,  because  5  one-sixteenths 
and  9  one-sixteenths  =  14  one-sixteenths  =  yfths.  Like  num- 
bers (see  Art.  8)  can  be  added,  but  unlike  numbers  cannot  be 
added;  thus.  4  feet  cannot  be  added  to  6  inches,  but  4  feet  can  be 
added  to  6  feet  and  4  inches  can  be  added  to  6  inches.  In  an 
expression  like  xV  +  tV,  the  denominators  show  what  is  to  be 
added,  in  this  case  16ths,  and  the  numerators  show  how  many  of 
the  things  indicated  by  the  denominator  are  to  be  added;  it  is 
exactly  the  same  operation  as  adding  5  dollars  and  9  dollars  and 
obtaining  14  dollars  for  the  sum.  In  one  case,  16ths  are  added, 
and  in  the  other  case  dollars  are  added.  Consequently,  if  the 
denominators  are  alike,  the  sum  of  several  fractions  may  be  found 
by  adding  the  numerators  and  writing  the  sum  over  the  denomi- 
nator.    For  instance,  A  +  iV  +  H  =  5  +  ^  13  =  U=  lH, 

the  last  result  being  obtained  by  dividing  the  numerator  by  the 
denominator. 

100.  If  the  denominators  are  unlike,  the  fractions  cannot  be 
added  until  the}*  have  been  reduced  to  a  common  denominator, 
preferably,  the  least  common  denominator;  thus,  §  cannot  be 
added  to  j,  because  2  one-thirds  +  4  one-fifths  has  no  particular 
meaning  as  it  stands — the  fractions  have  no  common  unit.  But, 
f  =  t!j  !  =  rfj  and  jt  +  tt  =  ff  =  Itt-  That  this  result  is 
correct  is  readily  seen.  Thus,  a  bushel  of  wheat  weighs  60 
pounds;  hence,  fds  of  a  bushel  weighs  40  pounds;  -fths  of  a 
bushel  weighs  48  pounds;  and  the  combined  weight  is  40  +  48 
=  88  pounds.  Now  T-7th  of  60  pounds  is  4  pounds,  and 
ffths  of  60  pounds  is  22  X  4  =  88  pounds,  the  same  result 
as  before. 

Similarly,  to  add  4  feet  and  6  inches,  it  is  necessary  to  express 
the  feet  in  inches  or  the  inches  in  feet  before  adding.  Since  there 
are  12  inches  in  1  foot,  there  are  12  X  4  =  48  inches  in  4  feet;  and 
48  imhcs  +  6  inches  =  54  inches.  If  it  is  desired  to  express  the 
sum  in  feet,  then  it  is  necessary  to  divide  54  by  12,  obtaining 
4 17  =  4|  feet.  The  same  result  might  have  been  obtained  by 
dividing  the  number  of  inches,  6,  by  12.  thus  obtaining  a  frac- 
tion of  a  foot,  which  can  then  be  added  to  the  number  of  feet; 
thus.  6  4-  12  =  A  =  h  that  is,  one-half  of  a  foot,  and  4  feet 
+  h  foot  =  4  +  £  =  4|  feet. 


§1  FRACTIONS  55 

101.  Rule. — I.  If  the  fractions  have  a  common  denominator,  add 
the  numerators  and  write  the  sum  over  the  denominator. 

II.  If  the  fractions  do  not  have  a  common  denominator,  reduce 
them  to  fractions  having  a  common  denominator,  preferably,  their 
least  common  denominator,  and  then  add. 

III.  //  the  sum  is  an  improper  fraction,  reduce  it  to  an  integer  or 
mixed  number  by  dividing  the  numerator  by  the  denominator. 

IV.  If  the  sum  is  a  proper  fraction  or  a  mixed  number  containing 
a  proper  fraction,  reduce  the  fraction  to  its  lowest  terms. 

Example. — Add  ,72,  \\,  JJ,  and  if. 

Solution. — Here  the  least  common  denominator  is  240.  Then,  Tv  +  }' 
+  «  +«  -  m  +  *85  +«M  +  m  -  140+165+204+130  _  639   '_ 

=  2£g.     Ans. 
In  practice,  the  work  would  be  arranged  as  follows: 

I     .    it    .    17    .    is  _   140  +  165  +  204+130  __     8, 

T5  +  Tff  +  2"o"  +  5T  —  240 — — "  —  Ans. 

Here  the  denominator  of  the  sum  is  written  only  once,  the 
numerators  of  the  fractions  being  wrii  ten  above  it  and  added. 
It  is  seen  at  a  glance  that  the  sum  of  the  numerators  is  greater 
than  the  denominator;  hence,  they  are  added  separately  and  the 
improper  fraction  reduced  to  a  mixed  number,  with  the  fractional 
part  in  its  lowest  terms. 

102.  The  Sum  of  Two  Fractions. — When  it  is  desired  to  add 
two  fractions  whose  denominators  are  unlike  and  have  no  common 
factor,  that  is,  when  these  are  prime  to  each  other,  proceed  as 
follows:  Add  yi  and  ^f.  Here  the  denominators  have  no  com- 
mon factor;  hence,  the  least  common  denominator  is  the  pro- 
duct of  the  denominators.  This  product  divided  by  either 
denominator  gives  the  other  denominator,  which  is  used  as  a 
multiplier  for  the  numerator.  Therefore,  multiply  the  denomina- 
tors for  a  new  denominator;  multiply  the  numerator  of  the  first 
fraction  by  the  denominator  of  the  second,  and  the  numerator 
of  the  second  fraction  by  the  denominator  of  the  first  and  add 
.,  ,  .  ,_,  n,  lt  11  X  24+ 19  X  13  264  +  247 
the  products.     Thus,  t£  +  27  = is  x  24^ =  — 3l2 — 

_  511 
~  312 


=  H?f.     Ans. 


Even  if  the  denominators  have  a  common  factor,  this  method  of 
adding  two  fractions  is  generally  to  be  preferred,  because  it  is 
quicker  than  reducing  the  fractions  to  their  least  common 
denominator. 


56  ARITHMETIC  §1 

103.  To  add  mixed  numbers,  add  the  integral  and  fractional 
parts  separately;  if  the  sum  of  the  fractions  is  an  improper  frac- 
tion, reduce  it  to  a  mixed  number  and  add  to  the  sum  of  the 
integers. 
Example.— What  is  the  sum  of  23f,  31H,  28M,  and  25|f  ? 
Solution*. — Arrange  the  numbers  in  the  same  manner  as  for  addition 
23.1     _     24  of    integers.     The  least  common  de- 

31|s  =    fl  nominator  of  the  fractions  is  evidently 

28H  =    r;5  64:  hence,  reduce  the  fractions  to  frac- 

o  rj4  8  _    4  3  ... 

8  ? El  tions   having   64  for  a  denominator, 

lUJftT        fi    —  ^5T  writing  them  as  shown.     Adding  the 

numerators,  the  sum  is  149,   and  the  sum  of  the  fractions  is  ^g-^-  =  2f|. 

Carry  the  2  into  the  column  of  integers,  which  add,  obtaining  109,  to  which 

annex  the  fraction,  obtaining  109fi  for  the  sum  of  the  mixed  numbers.     Ans. 

The  sum  might  have  been  obtained  by  reducing  the  mixed 
numbers  to  improper  fractions  and  then  adding  by  the  regular 
rule;  but  this  takes  longer  and  there  is  greater  liability  of  making 
a  mistake. 

As  another  example,  add  4f,  5|,  7f^,  and  6f.     The  least 
4I    =  iV<j  common  denominator  is  120.     Reducing 

7I1  Z  Ip  the  fractions  to  fractions  having  120  for  a 

6 1    =  -j^V  denominator,  the  sum  of  the  numerators 

24}i    Ans.  is  273,  which  divided  by  120  gives  a  quo- 

tient of  2{\-.  Carrying  the  2  to  the  sum  of  the  integers  and 
annexing  the  fraction,  the  sum  of  the  mixed  numbers  is  24-j-^; 
thus,  2  +  6  +  7  +  5  +  4  =  24,  and  24  +  ^  =  24^f 


EXAMPLES 

(1)  Add  f,  ft,  rV-  Ans.  1§J*. 

(2)  Add  \i  and  T%.  Ans.  l^V 

(3)  Add  i,  I,  |,  |,  A-  Ans.  \U. 

(4)  Add  Iff  and  §ff.  Ans.  lTV/sV. 

(5)  Add  HA,  8f,  9Jf,  10§.  Ans.  39fK 

(6)  Add  127$,  851,  109T\,  96f.  Ans.  419TVff. 

(7)  Three  kinds  of  dyes  were  added  to  a  beater  of  paper  stock,  and 
weighed  9%  ounces,  8^3  ounces  and  13%  ounces;  how  much  dye  stuff  was 
used?  Ans.  32}<j4  ounces. 


SUBTRACTION  OF  FRACTIONS 

104.  Since  subtraction  is  the  reverse  of  addition,  fractions  must 
be  reduced  to  a  common  denominator  before  the  subtraction  can 

6  —  5 
be  performed.     For  example,    f  —  f  =  — ~ —  =  |.      Here  the 


§1  TRACTIONS  57 

common  denominator  is  found  as  in  addition,  then  the  difference 
of  the  numerators  is  taken  and  the  remainder  is  written  over  the 
common  denominator. 

Example.— Find  the  difference  between  j£f  and  |y. 

Solution.— As  it  is  not  evident  which  fraction  is  the  larger,  assume  that 
the  first  fraction  is  the  larger;  if,  later,  it  is  found  not  to  be  such,  the  subtra- 
hend and  minuend  in  the  numerator  can  be  transposed.  Then,  proceeding 
as  in  addition  of  two  fractions, 

,„«  m  106X268-217X183  28408-28881  Tllis  oxpression 
hw-W--  ,33X268  58166  2g8M  _^o_8 

shows  that  the  fraction  |^  $  is  the  larger,  and  the  difference  is  -      .  s , . , . 

=  rUh-     Ans- 

105.  To  subtract  one  mixed  number  from  another,  subtract 
the  integral  and  fractional  parts  separately.     For  example,  sub- 
tract 26^  from  42f.     The  least  common  de- 
42§  =  J!  nominator  is  24;  hence,  reduce  the  fractions 

26 1  =  ft  to  fractions  having  24  for  a  denominator  and 

16ft     ft  subtract  as  shown.     The  difference  is  found 

to  be  16irV     A7is. 

Example. — From  109A  take  96  f. 

Solution.— Here  the  fraction  in  the  subtrahend  is  larger  than  the  fraction 

109^  =  H  l£f  =  U  in  the  minuend;  therefore,  proceed  exactly 

Q6^"  =  ||  ||         as  in  subtraction  of  integers  when  the  figure 

. in  the  subtrahend  denotes  a  larger  number 

12f|.     Ana.  than  the  figure  over  it  in  the  minuend,  by 

adding  1  to  the  fraction  in  the  minuend  and  1  to  the  integer  in  the  subtrahend 
Reducing  the  mixed  number,  1H  to  an  improper  fraction,  H  ~  *5 
=  ff,  and  109  -  97  =  12;  hence,  the  difference  is  12?f- 

In  a  similar  manner,  if  it  is  desired  to  subtract  a  fraction 
from  an  integer,  reduce  1  to  a  fraction  having  the  denominator 
of  the  given  fraction  and  subtract  1  from  the  integer.    Thus,  15 

106.  Rule.— .Reduce  the  fractions  to  fractions  having  a  common 
denominator,  find  the  difference  of  the  numerator*,  and  write  the 
remainder  over  the  common  denominator;  then  reduce  the  resulting 
fraction  to  its  lowest  terms. 


EXAMPLES 

(1)  Find  the  difference  between  H  and  Wi-  Ans 

(2)  From  11 A  subtract  7|. 

(3)  Subtract  8|  from  13.  A  "••  4^ 

(4)  From  472^  take  297  A  w-  ! '  5"V 


58                                     ARITHMETIC  §1 

(5)  What  number  added  to  53f  will  make  75£?  Ans.  21 f§. 

(6)  A  package  of  dye  weighs  3H  pounds;  after  1|  pounds  have  been 
used  how  much  remains?  Ans.  lj-f  pounds. 


MULTIPLICATION  OF  FRACTIONS 

107.  To  Find  the  Product   of   a   Fraction   and   an   Integer. 
In  Art.  53,  multiplication  was  shown  to  be  a  short  process  of 


addition;  hence,   4  X  f  may  be    regarded   as    f  +  f  +  t  +  t 


ue     legaiucu    as     T  ~T  T  ~r  T  ~l 

3+3+3+3      4X3       12 


•   „    =  It-     Here  it  is  seen  that  if  the 

numerator  of  the  fraction  be  multiplied  by  the  integer,  the  pro- 
duct is  the  same  as  the  product  of  the  integer  and  the  fraction. 
This  same  result  may  be  arrived  at  in  another  way,  thus:  fths  of 
4  is  evidently  3  times  ,th  of  4;  ,th  of  4  is  4  -r-  7  =  -f,  and  3 
times  4  one-sevenths  is  3  X  4  one-sevenths,  or  12  one-sevenths 
=  -V2-  =  ly,  as  before. 

Again,  what  is  the  product  of  5  and  yy?  Here,  5  X  tt 
=  it  =  isL=  3§.  This  same  result  may  be  obtained  by  can- 
celation; thus,  P  X  y~-  =  "V"  =  3f .     As  the  result  of  this  opera- 

3 

tion,  it  is  seen  that  dividing  the  denominator  of  the  fraction  by 
the  integer,  multiplies  the  fraction.  This  should  be  evident, 
since  the  smaller  the  denominator  the  larger  the  fraction;  and  if 
the  denominator  is  5  times  as  small,  the  fraction  must  be  5  times 
as  large.  Therefore,  the  product  of  a  fraction  and  an  integer  may 
be  obtained  by  multiplying  the  numerator  of  the  fraction  by  the 
integer  or  by  dividing  the  denominator  of  the  fraction  by  the 
integer.  If  the  denominator  is  not  a  multiple  of  the  integer,  but 
contains  one  or  more  factors  common  to  the  integer,  these  factors 
may  be  canceled  before  multiplying  the  numerator.  For  example, 
7 

iga  v  61       7X61       427        inft3 

4 

108.  To  Multiply  a  Fraction  by  a  Fraction. — When  the  word 

"of"  occurs  between  two  fractions  or  between  a  fraction  and  an 

integer,  it  has  the  same  meaning  as  the  sign  of  multiplication; 

thus,  f  of  |  and  f  of  12  have  the  same  meaning  as  |  X  f  and 


§1  FRACTIONS  59 

|  X  12,  respectively.     Now  f  is  2  times  J ;  3  of  \  is  V*>  since  1XJ 

is  I,  and  3  of  §  must  be  3  times  smaller,  o  v  o  =  2V     Since  |  is 

5  times  i,  £  of  f  must  be  5  times  as  large  as  |  of  |;  hence,  I  of 
f  =  iri  X  5  =  Tt.     Since,  also,  f  is  2  times  \,  and  3  of  f  is  A, 

!  of  f  is  A  X  2  =  H  =  A,  or  |j  X  %  =  A- 

12 

This  same  result  may  be  obtained  in  an  easier  manner  by 

multiplying  the  numerators  of  the  fractions  for  the  numerator 

of  the  product,  and  multiplying  the  denominators  of  the  fractions 

2x5 
for    the    denominator    of    the    product;    thus,    |  X  f  =  o  v  o 

4 
The  product  of  any  number  of  fractions  may  be  found  in  this 
same  way;  that  is,  divide  the  product  of  the  numerators  by  the 

% 

%    $     7     ;? 

product    of    the  denominators.     For  instance,  j,  X  s  X  7.  X  :. 

4        I 
=  A-     Cancelation  should  be  employed  whenever  possible. 

109.  To  multiply  a  mixed  number  by  an  integer,  multiply  the 
integral  and  fractional  parts  separately.     For  example,  17|  X  42 

7      21       147 
=  17  X  42  +  I  X  42;  ~  X  tf  =  -j-  =  36f ;    17  X  42  -  714; 

4 
and  714  +  36f  =  750|. 

110.  To  multiply  a  mixed  number  by  a  mixed  number,  reduce 
both  to  improper  fractions,  and  then  multiply  in  the  ordinary 

47       25 
94      17')       1 1 75 
manner.     Thus,  13?  X  10jf  =  ^  X  j£  =  -g-  =  146J- 

8 

111.  Rule. — I.  The  product  of  two  or  more  fractions  is  found  by 
dividing  the  product  of  the  numerators  by  the  product  of  the  denomina- 
tors, canceling  factors  that  are  common  to  the  numerators  and  denomi- 
nators whenever  possible  before  multiplying. 

II.  To  find  the  product  of  two  or  more  mixed  numbers,  reduce 
them  to  improper  fractions  before  multiplying. 


60  ARITHMETIC  §1 

EXAMPLES 

(1)  §  of  f  of  s  =  ?  Ans.  %%. 

(2)  H  X  H  -  T  Ans.  A. 

(3)  68  X  198  =  ?  4ns.  1348|. 

(4)  33H  X  56*  =  ?  4ns.  1894f|. 

(5)  3|  X  6f  X  A  =  ?  4ns.  llf. 

(6)  Htxm-  *  Ans-  "<;■*■ 

(7)  From  a  barrel  of  rosin  weighing  404  pounds,  Hth  of  the  contents 
were  taken;  how  many  pounds  were  taken?  Ans.  277 'f  pounds. 

(8)  A  car  load  of  coal  weighing  78,500  pounds  was  received.  At  the  end 
of  a  certain  period,  fths  had  been  used;  during  a  second  period,  one-half 
of  the  remainder  was  used;  and  during  a  third  period,  fths  of  what  was  left 
at  the  end  of  the  second  period  was  burned.  How  many  pounds  were  used 
during  the  three  periods?  Ans.  75,433Jf  pounds. 

(9)  Referring  to  the  last  example,  how  many  pounds  were  consumed 
during  each  of  the  three  periods?  f  First     period,      29437|     pounds. 

Ans.    j  Second    period,    24531 |    pounds. 
I  Third    period,    21464f£    pounds. 

(10)  What  is  the  value  of  4|  ounces  of  dye  that  is  worth  $1  ^  per  ounce? 

Ans.  $5H. 

(11)  From  a  pile  of  11,670  cords  of  wood,  %ths  was  used  in  one  part  of  a 
plant;  another  part  of  the  plant  used  %  as  much  as  the  first  part,  and  a  third 
part  of  the  plant  used  K2ths  as  much  as  the  second  part;  how  many  cords 
were  used  in  the  third  part?  Ans.  12761%2  cords. 

(12)  Referring  to  the  last  example,  how  many  cords  were  used  in  the 
first  and  second  parts  of  the  plant?         .  (  First     part,    4376K    cords. 

I  Second  part,  2188^  cords. 

Note. — Observe  that  the  product  of  two  proper  fractions  is  always  less  than  either 
of  the  fractions  used  in  finding  the  product. 


DIVISION  OF  FRACTIONS 

112.  To  Divide  a  Fraction  by  an  Integer. — Suppose  it  is  desired 
to  divide  -?  by  3.  Since  f  is  6  one-sevenths,  6  one-sevenths 
divided  by  3  is  2  one-sevenths  =  f,  in  the  same  way  that  6 
bushels  divided  by  3  is  2  bushels.  Hence,  dividing  the  numerator 
by  the  integer  divides  the  fraction.  Again,  the  quotient  of  f 
divided  by  3  is  evidently  only  one-third  as  much  as  the  quotient 
of  5  divided  by  1,  since  3  is  3  times  1;  hence,  the  quotient  of  " 
divided  by  3  will  be  §  of  f  =  |  X  f  =  A  =  ?•  Note  that  in 
the  first  case,  the  numerator  is  divided  by  the  integer,  while  in 
the  second  case,  the  denominator  is  multiplied  by  the  integer, 


§1  FRACTIONS  01 

the  result  being  the  same  in  both  cases.  Therefore,  dividing 
the  numerator  or  multiplying  the  denominator  by  an  integer 
divides  the  fraction. 

113.  To  Divide  an  Integer  by  a  Fraction. — Suppose  it  is  desired 
to  divide  12  by  f ;  for  instance,  how  many  boxes  holding  f 
pound  each  can  be  filled  from  a  package  holding  12  pounds?  If 
12  be  divided  by  1,  the  quotient  is  12,  but  if  12  be  divided  by  a 
number  smaller  than  1,  the  quotient  will  evidently  be  greater 
than  12.  Since  1  is  4  times  as  large  as  j,  12  divided  by  \  must 
be  4  times  as  large  as  12  divided  by  1;  hence,  12  4-  \  =  12  X  4 
=  48.  Since  f  is  3  times  \,  12  divided  by  f  will  be  only  one- 
third  as  much  as  12  divided  by  |,  and  the  quotient  will  be  48 
X  |  =  16,  or  48  4-  3  =  16  =  12  4-  f. 

This  same  result  may  be  obtained  in  a  much  easier  manner  by 
turning  the  fraction  upside  down  (this  is  called  inverting  the 

4       4 
fraction)    and  multiplying;  thus   }'l  X  ;J    =  16.     Consequently, 

to  divide  an  integer  by  a  fraction,  invert  the  fraction  and  multiply 
the  integer  by  the  inverted  fraction. 

114.  To  Divide  a  Fraction  by  a  Fraction. — Suppose  it  is  desired 
to  divide  f  by  f .  Since  the  divisor  is  a  fraction,  invert  it  and 
multiply  f  by  the  inverted  fraction;  thus,  f  X  I  =  I  =  l\. 
That  this  is  correct  is  easily  shown.  Since  f  is  3  times  \,  3  4-  § 
=  3X1  =  1,  and  one-fourth  of  £  is  f  4-  4  =  §■,  or  i  X  t  =  1 

=  li 

Example. — Divide  ,Y'>  by  $|. 

17 

7'J 
Solution. — Inverting    the    divisor    and   multiplying,    n.,    X    js    =   \\ 

1  I  9 

=   1,',.     Ans. 

115.  To  divide  a  mixed  number  by  an  integer,  divide  the  inte- 
gral part  by  the  integer;  if  there  is  a  remainder,  annex  the  fraction 

to  it,  reduce  this  new  mixed  number  to  an  improper  fraction,  and 
divide  it  by  the  integer. 

Ex  ample.— Divide  508|  by  8. 

Solution.— Here  508  -5-  8  =  63  and  4  remainder;  annexing  the  J  to  4, 
it  becomes  4f  =  ¥ ;  y  -j-8  =  ft.     Therefore,  508$  +-8=6 

The  mixed  number  might  have  been  reduced  to  an  improper 
fraction  and  then  divided  by  the  integer;  the  process  \wn-  given  is 
better,  however  (particularly,  when  the  divisor  is  small),  and  it 
requires  less  work. 


62  ARITHMETIC  §1 

116.  To  divide  a  mixed  number  by  a  mixed  number,  reduce 
both  to  improper  fractions,  and  then  proceed  in  the  regular 
manner. 

Example. — Divide  15|  by  3&. 

41        5 

Solution.— 15|   =  *§*;  3&  =  U;  ^  X  j|  =  W  -  4H-     ^««- 

2        21 

117.  Rule. — I.  To  divide  a  fraction  by  an  integer,  divide  the 
numerator  or  multiply  the  denominator  by  the  integer. 

II.  To  divide  an  integer  or  fraction  by  a  fraction,  invert  the 
divisor  and  multiply  the  integer  or  fraction  by  it. 

III.  To  divide  a  mixed  number  by  an  integer,  divide  the  integral 
part  by  the  divisor;  if  there  is  a  remainder,  annex  the  fractional  part 
to  it,  reduce  the  resulting  mixed  number  to  an  improper  fraction, 
and  divide  it  by  the  divisor;  if  there  is  no  remainder,  divide  the  frac- 
tional part  of  the  mixed  number  by  the  divisor.  The  sum  of  the  two 
quotients  is  the  entire  quotient. 

Example.— Divide  1296  f  by  16. 

2         11 
Solution.— 1296  4-  16   =  81;    1   -h  16  =  ^  X  jg   =  ^    hence,    1296f 

-5-  16  =  812**.  Ans.  That  this  method  of  dividing  a  fraction  by  an  integer 
is  correct  is  easily  seen.  Since  16  =  V,  inverting  the  divisor  makes  it 
Ts,  and  multiplying  by  ■£%  multiplies  the  denominator  of  the  fraction  by  16. 
This  method  of  dividing  by  an  integer  is  recommended,  because  it  permits 
of  cancelation,  when  possible. 


EXAMPLES 

(1)  Divide  |gf  by  12.  Ans.  ft. 

(2)  Divide  fi  by  4f.  Ans.  &. 

(3)  Divide  fjg  by  H-  &**■  2t- 

(4)  Divide  180  by  ||.  Ans.  328 

(5)  Divide  ^  by  \.  Ans.  \. 

(6)  Divide  19|  by  4J.  Ans.  4^. 

(7)  Divide  47951  by  21.  Ans.  228f. 


COMPLEX  AND  COMPOUND  FRACTIONS 

118.  A  complex  fraction  is  one  that  has  a  fraction  for  its  nu- 

3   2, 

8    3 
merator  or  denominator  or  both  terms  are  fractions;  thus,  ■='  ~ 


§1  FRACTIONS  63 

3 

g 
and  ^  are  complex  fractions.   In  order  to  distinguish  the  numerator 

5 

from  the  denominator,  the  dividing  line  is  made  heavier  than 
the  dividing  line  of  the  fraction  or  fractions.  Thus,  in  the  first 
complex  fraction,  the  numerator  is  f  and  the  denominator  is  7, 
the  expression  meaning  also  f  ■*■  7'}  in  the  second  complex  frac- 
tion, the  numerator  is  9  and  the  denominator  is  f ,  the  expression 
also  denoting  9  -J-  f ;  in  the  third  complex  fraction,  the  numerator 
is  f  and  the  denominator  is  f ,  the  expression  also  denoting  f  -f-  f . 

119.  To  simplify  a  complex  fraction,  multiply  the  digits  or 
numbers  above  and  below  the  heavy  line  by  the  denominator  of 
the  fraction;  the  product  will  be  the  denominator  of  the  simplified 
fraction  in  the  first  and  third  cases  above,  and  the  numerator 
in  the  second  case.  Or,  transpose  the  denominator  of  the  frac- 
tion from  above  to  below  the  line  or  from  below  to  above  the  line,as 
the  case  may  be,  and  use  it  as  a  multiplier.     For  example,  to 

3 
simplify   the   first    of   the    above    complex    fractions,    write  8 

7 

3  3     .      ,  ,  ^  = — = —  =  15;  in  the  third 

-  =  ts',  m  the  second  case,  o  S 


7X8 


5 


case,  o  =  q  v  o  =  f-     That  these  results  are  correct  may  be 

5 
proved  by  performing  the  operations  indicated.     Thus,   §4-7 

_         "  _       3    .     q      .      S     _   9X5  ,      3  s  5 

"7X8  ~  p'  J  ~  ^S       ~       '  s  —  t  -   s  X  ^r 

3X5  _  g 
3X8       8* 

Either  term  or  both  terms  of  a  complex  fraction  may  consist  of 
two  or  more  fractions  connected  by  one  or  more  of  the  four  signs 

11  _3 
1 1\      v 
of  operation,  +,  — ,  X,  -r- ;  thus,  — r —  is  a  complex  fraction, 

J29    • 

112        29 
and  it  is  equal  to  —r-  =  jvo- 


64  ARITHMETIC  §1 

120.  A  compound  fraction  is  a  fraction  of  a  fraction.  Thus, 
f  of  -it,  or  $  X  T8T>  is  a  compound  fraction.  Since  division  of 
a  fraction  by  a  fraction  is  changed  into  multiplication  by  in- 
verting the  divisor,  f  -4-  tt  may  also  be  called  a  compound 
fraction;  and  a  compound  fraction  may  be  denned  as  an  expres- 
sion containing  two  or  more  fractions  connected  by  signs  of 
multiplication  or  division.  Even  the  product  of  one  or  more 
integers  and  one  or  more  fractions  may  be  called  a  compound 
fraction;  thus,  145  X  fl,  |  of  4  of  147  =  \  X  f  X  147,  3|  X  288 
X  iV,  etc.  may  be  called  compound  fractions. 

121.  If  either  or  both  of  the  terms  of  a  complex  fraction  is  a 
compound  fraction,  the  best  way  to  simplify  it  is  to  use  the  method 

,      .      ,.,    160Xr2X  loooo  X 900X225 
of  Art.  119.     For  example,  simplify  —  — 33000 — 

4  714        2  9 

Transposing  the  two  denominators  -  -^  x  ^m  x  m^ 

7s      m     n 

100 

38X714X9       0,.122       . 
= Jqq =  2441f f .    Ans. 

First  cancel  the  ciphers  that  are  common  to  the  numerator  and 

denominator;  this  is  the  same  as  dividing  by  10,  100,  etc.     The 

cancelation  might  have  been  carried  farther,  since  38  and  714 

are  both  multiples  of  2,  and  100  is  a  multiple  of  4;  but  it  is  easier 

to  divide  by  100  than  by  25,  and  the  work  was  left  as  it  stands. 

29| 

As  another  example,  find  the  value  of  -^a       Q1  •     Reducing 

. o  8  X  oj 

the  mixed  numbers  to  improper  fractions, 

4 
4*  179XHX7  _5012  . 


X  v  43  X  3 


1 1  2  8 


Ans. 


INVOLUTION 

122.  The  product  of  several  equal  factors  is  called  a  power  of 
the  number  used  as  a  factor.  Powers  are  named  first,  second, 
third,  fourth,  fifth,  etc.  according  to  the  number  of  equal  factors 
considered  or  used.  Thus,  the  second  power  of  7  is  7  X  7  =  49, 
the  third  power  of  9  is  9  X  9  X  9  =  729,  the  fifth  power  of  6  is 
6X6X6X6X6  =  7776. 


§1  INVOLUTION  65 

123.  Instead  of  indicating  the  product  of  the  factors  as  above, 
it  is  customary  to  shorten  the  work  by  writing  a  .small  figure  above 
and  to  the  right  of  the  number  used  as  a  factor,  the  small  figure 
indicating  the  number  of  times  the  factor  is  to  be  used  and  cor- 
responds with  the  name  of  the  power.  Thus,  65  is  read  6  fifth  or 
6  to  the  fifth  power,  and  means  the  fifth  power  of  6,  or  6  X  6  X  6 
X  6  X  6  =  7776;  54  is  read  5  fourth  or  5  to  the  fourth  power,  and 
represents  5X5X5X5  =  625,  etc. 

The  second  and  third  powers  are  usually  called  the  square  and 
cube,  respectively;  thus,  21 2  is  read  21  square  and  equals  21  X  21 
=  441,  163  is  read  16  cube  and  equals  16  X  16  X  16  =  4096. 
The  first  power  of  any  number  is  the  number  itself;  thus  2171 
=  217,  481  =  48,  etc. 

The  small  figure  that  is  written  above  and  to  the  right  of  the 
number  is  called  an  exponent.  The  power  itself  is  the  product 
found  by  performing  the  multiplication;  thus,  the  cube  of  16  is 
4096,  the  square  of  21  is  441,  the  fifth  power  of  6  is  7776,  the  first 
power  of  528  is  528,  etc. 

Involution  treats  of  powers  of  numbers. 

124.  To  indicate  the  power  of  a  fraction,  enclose  the  fraction 
in  parenthesis  and  write  the  exponent  outside  the  parenthesis; 
thus,  the  cube  of  f  is  indicated  by  (f)3,  and  it  equals  f  X  | 

X  I  =    A  ...  . ,  A   =  II.     Hence,   to   raise   a  fraction  to  any 
4         4  X  4  X  4 

indicated  power,  raise  the  numerator  and  denominator  separately 

to  the  power  indicated.     For  example,  (A)4  = 

9X9X9X9  7461    .  ,        ,M4       94 

16  X  16  X  16  X  16  =  65536  m  °ther  WOrds*     (t(0    =  W 

U  9      v       9       V   — 9      V    — 9  =    — 

because  yz  X  tt  X  t'b"  X  i  o  -  ig  y  1  fi  y  16  X  16  ~  164 

126.  To  raise  10  to  any  power  indicated  by  the  exponent, 
simply  annex  to  1  as  many  ciphers  as  there  are  units  indicated  by 
the  exponent;  thus,  105  =  100,000,  103=  1000,  107  =  10,000,000, 
etc.  In  the  first  case,  the  exponent  5  indicates  5  units  and  5 
ciphers  follow  1 ;  in  the  second  case,  3  ciphers  follow  1  because  the 
exponent  is  3;  in  the  third  case,  7  ciphers  follow  1  because  the 
exponent  is  7,  etc.  These  results  may  all  be  proved  by  actual 
multiplication. 

Conversely,  to  express  any  power  of  10  as  10  with  an  exponent, 
count  the  number  of  ciphers,  and  the  number  so  found  will  be 
the  exponent.     Thus,  100  =  102;  10000  =  104;  1000000  =  106;  etc. 


66  ARITHMETIC  §1 

To  raise  a  mixed  number  to  the  power  indicated  by  the  ex- 
ponent, first  reduce  the  mixed  number  to  an  improper  fraction, 
and  then  raise  the  fraction  to  the  indicated  power.  Thus, 
(4A)2  =  (f!)2  =  -VtV-  =  22AV     Arts. 


EXAMPLES 

(2)  7462  =  ? 

Ans.  556,516. 

(2)  873  =  ? 

Ans.  658,503. 

(3)   (fi)3  =  ? 

1   ,,  .           9  26  1 

(4)  100,000,000  is  what  power  of  10? 

Ans.  The  8th. 

(5)   104  =  ? 

Ans.  10000. 

(6)   (29f)=  =  ? 

Ans.  877£i 

126.  To  Multiply  or  Divide  by  a  Power  of  10. — Consider 
what  occurs  when  some  integer,  as  7035,  is  multiplied  by  some 
power  of  10,  say  10,000.  According  to  Art.  58,  the  product 
is  found  by  multiplying  by  1  and  then  annexing  4  ciphers, 
the  number  of  ciphers  to  the  right  of  1;  thus,  7035 
This  operation  is  equivalent    to    the    following:    to  1000° 

multiply  any  number  (integer)  by  a  power  of  10,  70350000. 
simply  annex  to  the  number  as  many  ciphers  as  there  are  ciphers 
in  the  given  power  of  10  or  as  many  ciphers  as  are  indicated  by 
the  exponent  of  10.  For  instance,  10,000  =  104;  hence,  to 
multiply  any  number  by  10,000  or  104,  annex  4  ciphers  to  the 
integer. 

Now  any  integer,  as  7035,  may  be  written  7035.,  the  decimal 
point  being  always  understood  to  follow  the  unit  figure  (5,  in  this 
case)  whether  written  or  not;  but  when  it  is  written,  any  number 
of  ciphers  may  be  annexed  to  the  number  without  altering  its 
value.  Thus,  7035.0,  7035.0000,  and  7035  all  have  the  same 
value,  since  the  position  of  the  unit  figure  has  not  been  changed 
and  the  addition  of  the  ciphers  has  not  added  anything  to  the 
number.  If,  now,  7035.0000  be  multiplied  by  10,000,  the  product 
will  be  the  same  as  before,  or  70350000.,  the  decimal  point  being 
moved  4  places  to  the  right.  In  the  product,  5  no  longer  indicates 
5  units,  but  5  ten-thousands.  From  this,  it  is  evident  that  any 
number  may  be  multiplied  by  a  power  of  10  by  moving  (shifting) 
the  decimal  point  as  many  places  to  the  right  as  there  are  ciphers 
in  the  power  or  indicated  by  the  exponent  of  10.  Thus,  3.1416 
X  102  =  3.1416  X  100  =  314.16;    3.1416  X  104  =  3.1416  X 


§1  DECIMALS  AND  DECIMAL  FRACTIONS  07 

10000     =     31416;     3.1416     X     107    =    3.1416     X     10000000 
=  31416000;  etc. 

Rule. — To  multiply  any  number  by  a  -power  of  10,  shift  the 
decimal  point  as  many  places  to  the  right  as  there  are  ciphers  in  the 
power  or  indicated  by  the  exponent  of  10,  annexing  ciphers,  if 
necessary. 

127.  Since  3.1416  X  1000  =  3141.6,  3141.6  ■*■  1000  must  evi- 
dently equal  3.1416;  in  other  words,  to  divide  by  a  power  of  10, 
shift  the  decimal  point  as  many  places  to  the  left  as  there  are 
ciphers  in  the  power  or  indicated  by  the  exponent  of  10.  This 
evidently  follows  also  from  the  fact  that  division  is  the  reverse  of 
multiplication ;  hence,  if  the  decimal  point  is  shifted  to  the  right 
for  multiplication,  it  must  be  shifted  to  the  left  for  division. 

To  divide  3.1416  b}r  10,000,  it  is  necessary  to  prefix  ciphers  to 
the  first  figure,  3,  in  order  to  make  it  occupy  a  position  4  places 
to  the  right  of  unit's  place,  and  3.1416  •£■  104  =  3.1416  -r  10000 
=  .00031416.  In  multiplication,  the  unit  figure  is  removed  to 
the  left;  in  division,  it  is  removed  to  the  right.  Also,  7035  ■*■  107 
=  7035  •*■  10000000  =  .0007035. 

Rule.— To  divide  any  number  by  a  power  of  10,  shift  the  decimal 
point  as  many  places  to  the  left  as  there  are  ciphers  in  the  power  or 
indicated  by  the  exponent,  prefixing  ciphers,  if  necessary. 

It  will  be  noted  that  the  3  in  3.1416  indicates  3  units,  while  in 
.00031416,  it  indicates  3  ten-thousandths,  a  number  ten  thou- 
sand times  smaller  than  3  units.  This  is  evidently  correct,  since 
the  divisor  is  10,000. 


DECIMALS  AND  DECIMAL  FRACTIONS 


DEFINITIONS  AND  EXPLANATIONS 

128.  A  decimal  fraction  is  one  that  has  a  power  of  10  for  its 
denominator;  TV,  tWoV  3TVoW,  etc.  are  decimal  fractions. 
When  the  numerator  is  divided  by  the  denominator,  the  above 
fractions  become  .7,  .5236,  3.1416,  and  are  called  decimals. 

129.  Any  decimal  may  be  converted  into  a  decimal  fraction  by 
writing  the  decimal  (considered  as  an  integer)  for  the  numerator 
of  the  fraction  and  writing  for  the  denominator  a  power  of  10 
containing   as  many  ciphers  as  there  are  decimal  places  in  the 


68  ARITHMETIC  §1 

decimal.  For  example,  .7854  =  iVWv,  -00034  =  TvHinr,  .052465 
=  rot  off o,  etc. 

130.  The  value  of  a  decimal  is  not  changed  by  annexing 
ciphers;  thus,  .25000  =  .25.  This  is  evident,  since  nothing  has 
been  added  to  the  number  represented  bj-  the  decimal,  and  the 
position  of  the  first  digit  relative  to  the  decimal  point  has  not 
been  changed.     This  may  also  be  shown  by  converting  both 

25000 
decimals  into  decimal  fractions;  thus,  .25000  =  tt^Tw^a  =  iW 

and  .25  =  rVV  Here  it  is  seen  that  as  many  ciphers  are  added  to 
the  denominator  as  are  added  to  the  numerator,  and  these  may 
be  canceled  in  the  fraction. 

131.  If  the  numerator  of  a  decimal  fraction  contains  the  factor 
2  or  5,  the  decimal  fraction  may  be  reduced  to  lower  terms,  thus 
becoming  an   ordinary  common  fraction.     For    example,    .625 

—  tVA  =  iri  =  T¥  =  I )   -032  =  Tlhror  =  ttsS   .2704  =  iVoVtf 

—  169.   0f„ 

—  rih»j  etc. 

If  the  decimal  does  not  contain  2  or  5  as  a  factor,  that  is,  if  it 
does  not  end  in  5  or  is  not  an  even  number,  the  decimal  fraction 
cannot  be  reduced,  because  since  2  X  5  =  10  and  they  are  both 
primes,  2  and  5  are  the  only  factors  in  any  power  of  10.  Thus, 
1000  =  103  =  10  X  10  X  10  =  2  X  5  X  2  X  5  X  2  X  5 
=  2X2X2X5X5X5  =  23X53. 

132.  Addition  and  Subtraction  of  Decimals. — It  has  already 
been  shown  how  to  add  decimals  and  numbers  containing 
decimals;  simply  place  the  decimal  points  under  one  another,  and 
add  or  subtract  as  in  the  case  of  integers,  placing  the  decimal 
point  in  the  result  directly  under  the  decimal  point  in  the  num- 
bers added  or  subtracted.  The  only  case  that  need  be  considered 
here  is  that  in  which  the  subtrahend  contains  more  decimal 
places  than  the  minuend.  In  such  a  case,  annex  ciphers  to  the 
minuend  until  it  contains  as  many  decimal  places  as  the  subtra- 
hend, and  then  subtract. 

Example  1.— From  426.45  take  294.0847. 

Solution. — Here  the  minuend  contains  two  decimal  places  and  the  sub- 
trahend  contains   four;   therefore,    annexing   two   ciphers, 
426  4500  which  does  not  change  the  value  of  the  minuend,  the  sub- 

294  0847  traction  is  performed  as  in  the  case  of  integers.     This  opera- 

~-^r  tion  is  equivalent  to  reducing  the  decimal  fractions  to  a 

common  denominator,  which  in  this  case  is   10.000.     In 
practice,   the  ciphers  would  not  be  written;  they  would 
simply  be  considered  to  be  added. 


§1           DECIMALS  AND  DECIMAL  FRACTIONS  69 

Example  2.— From  93  take  77.5G52. 

Solution. — The  work  is  arranged  as  shown  in  the  margin  go 

with  the  decimal  points  under  each  other,  and  the  ciphers  --,  5652 

that  are  supposed  to  be  annexed  are  not  written,  but  are  -    ■ 

J.,  15.4348  Ans. 
understood. 


MULTIPLICATION  OF  DECIMALS 

133.  The  only  difference  between  multiplying  decimals  and 
multiplying  integers  is  the  locating  of  the  decimal  point  in  the 
product.  The  number  of  decimal  places  in  the  product  is  equal 
to  the  sum  obtained  by  adding  to  the  number  of  decimal  places  in 
the  multiplicand  the  number  of  decimal  places  in  the  multiplier. 
To  understand  the  reason  for  this,  multiply  two  decimal  fractions. 
Thus,  .24  X  .637  =  ^fo  X  VWV  =  tVWW  =  -15288.  Here  it 
will  be  noted  that  the  number  of  ciphers  in  the  denominator  of 
the  product  is  equal  to  the  sum  of  the  number  of  ciphers  in  the 
denominators  of  the  two  fractions.  But  the  number  of  ciphers 
in  the  denominators  is  the  same  as  the  number  of  decimal  places 

637  in    the    factors;    therefore,     multiply     the 

.24  numbers  in  the  same  manner  as  though  they 

2548  were  integers,  disregarding  the  decimal  points. 

1274  The  product  is  found  to  be  15288.     There  are 

.15288  Ans.  3  decimal  places  in  the  multiplicand  and  2  in 

the  multiplier;  hence,  there  are  3  +  2  =  5  decimal  places  in 

the  product. 

Had  the  numbers  been  .037  and  .24,  the  pro'duct  as  integers 
would  have  been  888;  pointing  off  5  decimal  places,  the  product 
as  decimals  is  .00888. 

134.  Mixed  numbers  are  multiplied  in  exactly  the  same  way  as 
integers,  and  the  number  of  decimal  places  in  the  product  is 
determined  in  the  same  manner  as  in  multiplication  of  decimals. 

Example  1.— What  is  the  product  of  75.305  and  3.1416? 

7  5.305  Solution. — The  product  of  the  numbers  is  found 
3.1416  as  though  they  were  integers.  Since  there  are  3 
4  5  1  S  3  0  decimal  places  in  the  multiplicand  and  4  in  the 
7  5  3  0  5  multiplier,  7  decimal  places  are  pointed  off  in  the 
3  0  12  2  0  product,  which  is,  therefore,  236.578188. 

7  5  3  0  5 

2  2  5  9  15 

23  6. 5781880  Ans. 


70  ARITHMETIC                                    §1 

Example  2.— Find  the  product  of  47,082  and  .0005073. 

4  7  0  8  2  Solution. — The   numbers   are   multiplied   as  in 

.  0  0  0  5  0  7  3  example   1,   as    though    they    were    integers.     The 

14  12  4  6  multiplicand   contains  no    decimal  places  and  the 

3  2  9  5  7  4  multiplier  contains  7;  hence,  7  decimal  places  are 

2  3  5  4  10  0  pointed  off  in  the  product. 

2  3.8846968  Ans. 


EXAMPLES 

(1)  47.95  X  126.42  =  ?  Ans.  6061.839. 

(2)  .0903  X  .7854  =  ?  Ans.  .07092162. 

(3)  730.25  X  36.524  =  ?  Ans.  26671.651. 

(4)  1285  X  .0001016  =  ?  Ans.  .130556. 

(5)  .00575  X  .00036  -  ?  Ans.  .00000207. 

(6)  What  will  be  the  cost  of  7042  tons  of  coal  at  $7.33  per  ton? 

Ans.  $51,617.86. 

(7)  If  the  price  of  a  certain  dye  is  $1.13  per  ounce,  how  much  must  be 

paid  for  18.625  ounces?  Ans.  $210.4625,  or  $210.46. 

Note. — In  all  monetary  transactions,  if  the  fractional  part  of  a  cent  is  equal  to  or  greater 
than  J  the  number  of  cents  is  increased  by  1;  otherwise,  the  fraction  is  rejected.  In  the 
last  example,  the  fraction  of  a  cent  is  .25  (found  by  moving  the  decimal  point  so  as  to 
follow  the  cent),  and  it  is  therefore  rejected. 

(8)  What  amount  must  be  paid  for  5050  pounds  of  paper  at  12.375  cents 
per  pound?  Ans.  62,554  cents,  or  $625.54. 

(9)  How  much  freight  will  have  to  be  paid  on  439  tons  of  limestone  at 
$4.17  per  ton?  Ans.  $1830.63. 

(10)  If  a  pulp  mill  uses  an  average  of  9.07  tons  of  soda  ash  per  month 
and  the  cost  is  $45  per  ton,  how  much  will  be  paid  for  this  material  in  a  year? 

Ans.  $4897.80 


DIVISION  OF  DECIMALS 

135.  Decimals  are  divided  in  exactly  the  same  manner  as 
integers,  no  attention  being  paid  to  the  decimal  point  until  all 
the  figures  of  the  dividend  have  been  used.  To  understand  how 
the  decimal  point  is  located  in  the  quotient,  consider  carefully  the 
following  examples  and  their  solutions,  paying  strict  attention  to 
the  explanations. 

Example  1.— Divide  7.092162  by  .0903. 

Solution. — The  division  is  performed  as  shown  in  the 
margin,  the  quotient  being  7854  when  the  dividend  and 
divisor  are  considered  as  integers.  To  locate  the  decimal 
point,  subtract  the  number  of  decimal  places  in  the  divisor 
from  the  number  of  decimal  places  in  the  dividend,  and  the 
remainder  will  be  the  number  of  decimal  places  in  the  quo- 
tient. In  the  present  case,  the  divisor  contains  4  decimal 
places  and  the  dividend  6;  hence,  there  are  6—4  =  2 
decimal  places  in  the  quotient. 


7.092162(0903 

6  321. 

7S.54 

7711 

Ans. 

7224 

4876 

4515 

3612 

3612 

§1  DECIMALS  AND  DECIMAL  FRACTIONS  71 

To  understand  the  reason  for  above  proceeding,  convert  the 
decimals  into  decimal  fractions  and  divide  by  the  rule  for  division 


7092162  903_         7092162 

of  fractions,     ihus,    1000000     •     1C)000    :      10OPPPP    X     903 

=    1f)r>   =  78.54.     Here  4  ciphers  in  the  numerator  of  the  divisor 

cancel  4  ciphers  in  the  denominator  of  the  dividend,  leaving  6 
—  4  =  2  ciphers  in  the  denominator  of  the  quotient.  Since  the 
number  of  ciphers  in  the  denominators  of  the  fractions  is  the  same 
as  the  number  of  decimal  places  in  the  corresponding  numbers 
expressed  as  decimals,  the  number  of  decimal  places  in  the  quo- 
tient is  always  equal  to  the  number  of  decimal  places  in  the  divi- 
dend minus  the  number  in  the  divisor. 

Example  2.— Divide  7.092162  by  903. 

Solution. — Considering  the  dividend  as  an  integer,  the  quotient  is  7854, 
as  found  in  the  last  example.  Since  there  are  no  decimal  places  in  the 
divisor  and  there  are  6  in  the  dividend,  there  are  6—0  =  6  decimal  places 
in  the  quotient,  which  is  therefore  equal  to  .007854.  That  this  is  correct 
may  be  proved  by  multiplying  the  quotient  by  the  divisor. 

Example  3.— Divide  70921.62  by  .0903. 

Solution. — Considering  both  numbers  as  integers,  the  quotient  is  7854. 
Since  the  divisor  contains  a  greater  number  of  decimal  places  than  the  divi- 
dend, annex  ciphers  to  the  dividend  (this  does  not  change  its  value,  see  Art. 
130)  until  it  contains  the  same  number  as  the  divisor;  in  this  case,  2  ciphers, 
then  since  both  dividend  and  divisor  contain  the  same  number  of  decimal 
places,  in  this  case  4,  the  quotient  is  an  integer,  because  4—4=0,  and 
there  are  no  decimal  places  to  be  pointed  off.  Therefore,  70921.62 
-h  .0903  =  70921.6200  -i-  .0903  =  7S5400,  treating  the  dividend  and  divisor 
as  integers. 

The  same  result  may  also  be  arrived  at  as  follows :  Expressing 
the  decimals  as  decimal  fractions  and  dividing  as  in  example  1, 

^  X  1^  ^  X  100  =  7854  X  !00  =  785400. 

Or,  since  multiplying  both  numerator  and  denominator  by  the 
same  number  does  not  alter  the  value  of  the  fraction,  •1±iJiAi 
X  !##  =  -^Hlf^j  a  fraction  whose  denominator  is  the  same  as 
the  denominator  of  the  divisor,  and  which  equals  the  decimal 
70921.6200,  which,  in  turn,  is  the  dividend  with  2  ciphers  an- 
nexed. But,  — TOWO-  X  ^903~  =  785400'  the  same  result  as 
before. 


72 


ARITHMETIC 


§1 


4575.120(7.854 
3927  0 


648  12 
628  32 
19  800 
15  708 
4  0920 
3  9270 


582WV    Ana. 
Or,  582. 521-i-}  * 


Example  4.— Divide  4575.12  by  7.854. 

Solution. — Since  the  divisor  contains 
one  more  decimal  place  than  the  dividend, 
annex  a  cipher  to  the  dividend.  The 
quotient  is  found  to  be  582,  and  there 
is    a    remainder    of    4092;    hence,    the 

Ans-  quotient  is  5824ff|-  =  582^%,  a  mixed 
number  composed  of  an  integer  and  a 
common  fraction.  It  is  generally  more 
convenient  to  express  the  quotient  in 
such  cases  as  a  mixed  number  composed 
of  an  integer  and  a  decimal;  in  this  case, 
simply  annex  one  or  more  ciphers  to  the 
remainder  and  proceed  with  the  di- 
vision as  indicated.  In  other  words, 
^Y9  =  .521TY9-.  That  this  is  the  case 
is    readily    shown.     Thus,    62  =  62.000, 

?         and     62.000    -h    119   =  .521T^.     Note 

that  three  ciphers  in  all  were  annexed 

to  the  remainders,  since  there  were  three 

different  remainders  used  as  dividends,  and  one  cipher  was  annexed  to  each. 


40 !) 2    _ 


6  2 


R  fi 


348.0000000(25575 
255  75 


0136070  .4  ns. 


92  250 

76  725 

15  5250 

15  3450 
180000 
179025 
9750 


Solution. — Since  there  are  no  decimal 
places  in  either  dividend  or  divisor,  annex  as 
many  ciphers  to  the  dividend  following  the 
decimal  point  as  there  are  decimal  places 
required  in  the  quotient,  in  this  case  7. 
When  all  the  figures  in  the  dividend  have 
been  used,  there  will  be  7  decimal  places  in 
the  quotient,  because  7—0  =  7.  The 
quotient  to  7  decimal  places  is  .0136070. 


136.  Whenever  the  divisor  contains  a  factor  other  than  2  or  5, 
there  will  always  be  a  remainder,  no  matter  to  how  many  decimal 
places  the  quotient  is  carried,  and  as  this  is  usually  the  case,  it  is 
customary  to  carry  the  division  one  place  farther  than  the  num- 
ber of  decimal  places  specified.  If,  then  the  extra  figure  is  5  or  a 
greater  digit,  the  preceding  figure  is  increased  by  1,  and  a  minus 
sign  is  written  after  the  quotient  to  show  that  it  is  not  quite  so 
large  as  written  or  printed;  but,  if  the  extra  figure  is  less  than  5,  it 
is  rejected,  and  a  plus  sign  is  written  after  the  quotient  to  show 
that  it  is  a  little  larger  than  written  or  printed.  In  this  example, 
it  is  seen  that  the  figure  in  the  8th  decimal  place  is  3;  hence,  the 
quotient  to  7  decimal  places  would  be  written  .0136070  +.  The 
quotient  to  3  decimal  places  would  be  written  .014  — .  In  prac- 
tice, these  signs  may  be  omitted. 


§1  DECIMALS  AND  DECIMAL  FRACTIONS  73 

Example. — Find  the  quotient  of  .0348  +  .00255  to  five  decimal  places. 
.03480  ( . 00255 
255        13.647058 

930      13.64706—  Ans.      Solution. — The  divisor  contains  one  more 

7g5  decimal  place  than  the  dividend;  hence,  annex- 

TT70  ing  one  cipher  to  the  dividend,  the  integral  part 

1  -OQ  of  the  quotient  is  found  to  be  13.     Ciphers  are 

■ now  annexed  to  the  different  remainders  as  de- 
scribed in  example  4,  and  5  +  1=6  figures  of 

.  the  decimal  are  found.     Since  the  sixth  figure 

1800  of  the  decimal  is  8,  the  preceding  figure,  5,  is 

1785  increased  by  1,  and  the  quotient  to  five  decimal 

1500  places  is  13.64706-. 
1275 
2250 
2040 

137.  Rule — I.  //  the  divisor  does  not  contain  more  decimal 
places  than  the  dividend,  divide  as  though  the  numbers  were  integers; 
subtract  the  number  of  decimal  places  in  the  divisor  from  the  number 
in  the  dividend,  and  point  off  in  the  quotient  as  many  decimal  places 
as  are  indicated  by  the  remainder. 

II.  If  the  divisor  contains  a  greater  number  of  decimal  places  than 
the  dividend,  annex  ciphers  to  the  dividend  until  it  contains  the  same 
number  of  decimal  places  as  the  divisor;  if  the  dividend  {with  the 
annexed  ciphers)  is  then  larger  than  (he  divisor,  both  being  considered 
as  integers,  the  quotient  when  all  the  figures  of  the  dividend  (including 
the  annexed  ciphers)  have  been  used  will  be  an  integer.  But,  if  the 
dividend  is  smaller  than  the  divisor,  annex  ciphers  to  the  dividend 
and  proceed  with  the  division;  the  quotient  will  be  a  decimal,  and 
may  be  pointed  off  as  in  I. 

III.  //  there  be  a  remainder,  it  may  be  written  over  the  divisor, 
and  the  resulting  fraction  reduced  to  lower  terms,  if  possible.  Or, 
ciphers  may  be  annexed  to  the  remainders  as  found  and  the  division 
continued  until  the  quotient  has  been  determined  to  any  desired 
number  of  decimal  places. 


EXAMPLES 

(1)  Divide  268.92  by  .007007.  Ans.  38378.76  + 

(2)  Divide  304.75  by  3.125.  Ans.  97.52. 

(3)  Divide  100.43  by  1000.26.  Ans.  .100104-. 

(4)  Divide  648.433  by  6509.85.  A  ns.  .099608-. 

(5)  Divide  .00218  by  966.  Ans.  .000002257-. 


74  ARITHMETIC  §1 

(6)  The  price  of  a  certain  dye  is  $19.25  per  pound ;  if  the  bill  for  a  purchase 
of  dye  amounts  to  $1398.88,  how  many  pounds  were  bought? 

Ans.  72.67—  pounds. 

(7)  The  amount  paid  for  3490  tons  of  coal  was  $25,642.12;  what  was  the 
price  per  ton?  Ans.  $7.35. 

(8)  Bought  press  felts  at  $2.16  per  pound.     The  bill  came  to  $546.48; 
what  was  the  weight  of  the  felt?  Ans.  253  pounds. 


DECIMALS  AND  COMMON  FRACTIONS 

138.  To  reduce  a  common  fraction  to  a  decimal,  divide  the 

numerator  by  the  denominator.     If  the  denominator  is  a  power 

of  2,  as  2,  4,  8,  16,  32,  64,  etc.,  an  exact  decimal  equivalent  can  be 

found  for  the  common  fraction;  this  will  also  occur  when  the 

denominator  is  a  power  of  5,  as  5,  25,  125,  625,  3125,  etc.     But,  if 

the  denominator  contains  any  other  factor  than  2  or  5,  the  quotient 

will  never  be  exact,  no  matter  to  how  many  decimal  places  the 

division  may  be  carried.     In  such  cases,  find  the  quotient  to  one 

more  decimal  place  than  is  desired;  if  the  extra  figure  is  5  or  a 

greater  digit,  increase  the  preceding  figure  by  1  and  annex  the 

sign  —  ;  otherwise,  reject  it  and  annex  the  sign  +  . 

Examples. — Reduce  (a)  y£,  (6)  i|,  and  (c)  sf 7  to  decimal  fractions : 
Solution. — The   work   is   shown   herewith.     In    (a),   the   first  figure  of 


(a) 

(b) 

(c) 

11  0  (16 

17 

.0  (24 

5 

.000  (657 

9  6  .6875 

Ans. 

16 

.8  .70833+  Atis. 
200 

4 

599  .0076103+  Ans. 

1  40 

4010 

1  28 

192 

3942 

120 

To 

680 

112 

72 

657 

80 

80 

2300 

80 

72 

1971 

8 

329 

the  quotient  evidently  denotes  .6,  since  it  was  necessary  to  annex  one  cipher 
to  the  dividend,  11,  before  dividing;  and  since  the  divisor  (denominator)  is 
a  power  of  2  (24  =  16),  the  quotient  is  exact,  the  remaining  figures  being 
found  by  annexing  ciphers  to  the  different  remainders. 

In  (b),  the  first  figure  of  the  quotient  denotes  .7;  the  remaining  figures  are 
found  as  in  (a).  Note  that  the  second  remainder,  and  all  succeeding 
remainders,  is  8,  and  that  the  third  figure  of  the  quotient,  and  all  succeeding 
figures,  is  3. 

In  (c),  it  is  necessary  to  annex  3  ciphers  to  the  dividend  (numerator) 
before  it  will  contain  the  divisor   (denominator);  hence,  the  first  figure 


§1  DECIMALS  AND  DECIMAL  FRACTIONS  75 

of  the  quotient  denotes  .007.  The  division  is  here  carried  to  7  decimal 
places.  The  next  figure  will  be  5,  and  the  correct  value  of  the  quotient  to 
7  decimal  places  is  .0076104  —  . 

139.  Mixed  Fractions.— It  sometimes  happens  that  it  is 
desirable  to  express  a  quotient  exactly,  even  though  it  has  been 
reduced  to  an  approximate  decimal.  For  instance,  in  (6)  of  the 
last  example,  the  quotient  might  have  been  written  .7082Hf  = 
.708^,  and  the  quotient  in  (c)  might  have  been  written  .00761  b^V 
Expressions  of  this  kind  are  called  mixed  fractions. 

A  mixed  fraction  may  be  reduced  to  a  common  fraction  in  the 
same  way  that  a  mixed  number  is  reduced  to  an  improper 
fraction;  that  is,  multiply  the  decimal  by  the  denominator  of  the 
fraction,   add   the   numerator   to   the   product,   and   write   the 

....  .708  X  3  +  .001 
sum    over   the    denominator.     Thus,    .7O83    = ^ 

2. 124 +.001       2.125      _  ,      ,  ,,      ,      ..      , 

_  a.la-x  -r =        —     The  numorator,  1,  of  the  fraction  be- 

3  3 

longs  to  the  same  order  as  the  figure  of  the  decimal  that  precedes  it ; 

in  other  words,  .708|  =  iVo8o  +  Woo  =  -708  +  -g-     To  get  rid 

of  the  decimal  in  the  numerator  of  the  reduced  fraction,  note 

2  125  X  8 
that  .125  =  I j  hence,  multiply  both  terms  by  8,  and     3       g 

.00761  X  657  +  .00023 
=  H-     Again,  .00761  15-57  =  —        "657 


4.99977  +  .00023 


—  B5T' 


657 

In  practice,  the  work  would  be  performed  as 

shown   in   the   margin.     Here   the   decimal   is 

657  multiplied  by  the  denominator,  and  the  num- 

— ^^  erator  is  added  to  the  product  before  pointing 

3805  off,  both  the  product  and  the  numerator  being 

4  566  treated  as  integers. 

4  99977  140>  To  Reduce   a   Decimal  to   a   Fraction 

Havinc  a  Given  Denominator. — To  reduce  a 
decimal  to  a  fraction  having  a  specified  de- 
nominator, all  that  is  necessary  is  to  multiply 
the  decimal  by  the  specified  denominator;  the  product  will  be  the 
numerator  of  the  fraction.  Thus,  to  reduce  .671875  to  a  fraction 
having  64  for  its  denominator,  .671875  X  64  =  43;  hence,  the 
fraction  is  ?f . 


76  ARITHMETIC  §1 

The  reason  for  this  is  simple.      Xo  number  is  changed  by 

671875  X  64 
multiplying  it  by  1;  1  -  ft;  and  .671875  X  ft  =  : ^ 

=  *f.     Or,  reducing  the  decimal  to  a  decimal  fraction,  .671875 
=  tVoWo^;  now  multiplying  both  terms  of  this  fraction  bv  64, 

.„    ,    ,  .  671875X64       43000000  * 

the  specified  denominator,  moQOO  x  ^  =  W000000  =  ** 

In  the  case  just  given,  the  product  of  the  decimal  and  the 
specified  denominator  is  an  integer,  43.  This  seldom  occurs  in 
practice,  and  what  is  usually  sought  is  that  fraction  having  the 
specified  denominator  that  is  nearest  in  value  to  the  decimal. 
For  instance,  what  fraction  having  a  denominator  of  32  is  nearest 
in  value  to  .708?  Since  .708  X  32  =  22.656,  =  23-,  .708 
=  ||,  approximately.  Had  the  decimal  been  .703,  .703  X  32 
=  22.496  =  22  +,  and  .703  =  f|  =  ii,  approximately.  Here, 
as  before,  if  the  first  figure  of  the  decimal  part  of  the  product  is  5 
or  a  greater  digit,  increase  the  preceding  figure  by  1 ;  otherwise, 
reject  the  decimal.  Had  it  been  required  to  express  .703  as  a 
fraction  having  a  denominator  of  64,  .703  X  64  =  44.992 
=  45  — ,  and  .703  =  ?|,  approximately. 


EXAMPLES 

(1)  Reduce  §f|  to  a  decimal.  Ans.  .80859375. 

(2)  Reduce  .S035H  to  a  common  fraction.  Ans.  ftHflii. 

(3)  Reduce  T§ST  to  a  decimal.  Ans.  .0061597  —  . 

(4)  Reduce  Ti||5  to  a  mixed  decimal.  Ans.  .0569S36S4T. 

(5)  Express  .S0S6  inch  to  the  nearest  G4th  of  an  inch. 

Ans.  ||  =  ft  inch. 

(6)  Express  as  decimals  f,  f,  f.  Ans.  .375,  .625,  .875. 

(7)  Express  .3937  to  the  nearest  12th.  Ans.  &. 

(8)  Express  .3937  inch  to  the  nearest  32d  inch.  Ans.  £f  inch. 

(9)  Reduce  .05698|f  pound  to  a  common  fraction. 


SIGNS  OF  AGGREGATION 

141.  When  it  is  desired  to  indicate  that  several  numbers  are 
to  be  operated  upon  as  though  they  were  a  single  number,  a  sign 
of  aggregation  is  used  to  enclose  the  numbers.  The  word  aggre- 
gate means  to  collect,  to  bring  several  things  together.  There  are 
four  of  these  signs:  the  vinculum  ,  a  straight  line  placed  over 
the  numbers,  and  the  parenthesis  (    ),  brackets  [    ],  and  brace  {    }, 


§1  RATIO  AND  PROPORTION  77 

used  to  enclose  the  numbers.  An  expression  like  3  X  (45  —  28) 
or  3  X  45  -  28  means  that  28  is  to  be  subtracted  from  45  and  the 
remainder  is  to  be  multiplied  by  3.  In  arithmetic,  only  the 
vinculum  and  the  parenthesis  are  generally  used. 

142.  Order  of  Signs  of  Operation. — When  several  numbers  are 
connected  by  signs  of  operation  denoting  addition  and  subtrac- 
tion, perform  the  operations  indicated  in  regular  order  from  left 
to  right;  thus,  25  -  19  +  47  -  32  -  9  +  50  =  6  +  47  -  32 
-  9  +  50  =  53  -  32  -  9  +  50  =  21  -  9  +  50  =  12  +  50 
=  62.  If,  however,  a  sign  of  multiplication  or  division  occurs,  the 
operation  indicated  by  this  sign  must  be  performed  before  adding 
or  subtracting;  thus,  25  -3X5  + 42 -^7  + 38  =  25  -15 
+  6  +  38  =  10  +  6  +  38  =  16  +  38  =  54. 

If  signs  of  multiplication  and  division  follow  one  another  with 
no  signs  of  addition  or  subtraction  between  them,  perform  the 
operations  of  multiplication  and  division  in  order  from  left  to 
right  before  adding  or  subtracting;  thus,  46  +  54  -5-  6  X  8  -  67 
=  46  +  9  X  8  -  67  =  46  +  72  -  67  =  118  -  67  =  51. 

Now  by  using  signs  of  aggregation,  the  order  of  operations 
indicated  by  the  signs  of  operation  may  be  changed;  thus,  if 
desired,  the  last  expression  may  be  written  (46  +  54  -=-  6) 
X  8  -  67.  Here  54  is  first  divided  by  6,  the  quotient  is  added  to  46, 
the  sum  is  multiplied  by  8,  and  67  is  then  subtracted,  the  result 
being  (46  +  9)  X  8  -  67  =  55  X  8  -  67  =  440  -  67  =  373. 
When  one  sign  of  aggregation  includes  another,  as  in  this  case, 
always  consider  the  inner  sign  first. 


EXAMPLES 

(1)  159  -  (8  +  5  X  16)  +  11  -  100  -  ?  Ana.  51. 

(2)  6  X  13  -  119  -5-  7  -  50  -s-  12  =  ?  Ana.  56%. 

(3)  20  +  (126  -  4  X  270  -5-  9)  X  15  -  35  =  ?  Ana.  75. 

(4)  96  -  7.3  X  11  -5-  14  +  24  X  8  =  ?  Ana.  282.2ft. 


RATIO  AND  PROPORTION 

RATIO 

143.  It  is  frequently  desirable  to  ascertain  the  relative  sizes  of 
two  numbers.  For  instance,  suppose  it  is  desired  to  know  the 
relative  sizes  of  28  and  7,  that  is,  how  many  times  7  is  28  ?    Divid- 


78  ARITHMETIC  §1 

ing  28  by  7,  28  •*■  7  =  4;  hence,  28  is  4  times  7,  or,  in  other  words, 
28  is  4  times  as  large  as  7.  What  has  here  been  done  is  to  compare 
28  and  7,  the  comparison  being  done  by  division.  If  it  were 
desired  to  compare  7  to  28,  that  is,  to  find  what  part  of  28  is  7, 

divide  7  by  28,  obtaining  7  -f-  28  =  j;  in  other  words,  7  is  3>£th 

of  28  or  7  is  3^th  as  large  as  28. 

When  two  numbers  are  compared  in  this  manner,  by  dividing 
the  first  by  the  second,  the  comparison  is  called  a  ratio.  The 
language  used  in  the  first  of  the  above  cases  is  the  ratio  of  28  to  7, 
and  in  the  second  case,  the  ratio  of  7  to  28;  in  either  case,  the  ratio 
of  one  number  to  another  is  the  first  number  divided  by  the 
second. 

144.  A  ratio  may  be  indicated  as  above  by  using  the  sign  of 
division,  but  it  is  usual  to  use  the  colon  (see  Art.  64),  thus  indicat- 
ing distinctly  that  a  ratio  is  implied ;  thus,  the  ratio  of  57  to  36  is 
written  57  :  36.  A  ratio  is  also  very  frequently  indicated  in  the 
form  of  a  fraction;  thus,  the  ratio  of  128  to  16  is  written  128  :  16 

128 
or  -jjt.     The  second,  or  fractional,  form  possesses  many  advan- 

1Q 
12* 

The  two  numbers  used  in  forming  or  expressing  a  ratio  are 
called  the  terms  of  the  ratio.  The  number  to  the  left  of  the 
colon  or  above  the  dividing  line  is  called  the  first  term,  and  the 
other  number  is  called  the  second  term.     In  the  ratio  32  :  12, 

32 

or  yo*  32  is  the  first  term  and  12  is  the  second  term. 

The  value  of  a  ratio  is  the  quotient  obtained  by  dividing  the 
first  term  by  the  second  term;  thus,  the  values  of  the  ratios  57  : 

32 

36,  16  :  128,  ys,  etc.  are  l^,  h  2f,  etc.  respectively. 

145.  Ratios  like  the  above  in  which  the  first  term  is  named  first 
in  speaking  or  writing  are  called  direct  ratios.  It  is  frequently 
convenient  to  name  the  second  term  first,  in  which  case  the  ratio 
is  called  an  inverse  ratio ;  thus,  the  direct  ratio  of  24  to  4  is  24  :  4, 
and  its  value  is  6;  but  the  inverse  ratio  of  24  to  4  is  4  :  24,  and  its 
value  is  f .  The  word  direct  is  seldom  or  never  used  in  connection 
with  ratio;  but  if  the  ratio  is  an  inverse  one,  the  word  inverse  or 
inversely  is  always  used.  If  there  is  nothing  to  show  that  the 
ratio  is  inverse,  it  is  always  taken  to  be  direct. 


tages.     The  ratio  of  16  to  128  is  written  16  :  128  or  ,, 


§1  RATIO  AND  PROPORTION  79 

The  best  way  of  writing  an  inverse  ratio  is  to  write  it  first  as  if 
it  were  direct,  and  then  invert,  i.e.,  transpose,  the  terms.  For 
example,  to  write  the  inverse  ratio  of  36  to  90,  write  first  36  :  90 

or  q^;  now  invert  the   terms   and  obtain  90  :  36  or  ^z.     The 

value  of  the  inverse  ratio  of  36  to  90  is  2\  =  2.5. 

146.  Evidently,  only  like  numbers  can  be  used  to  make  up  the 
terms  of  a  ratio.  For  instance,  5  dollars  cannot  be  compared 
with  8  feet;  but  5  dollars  can  be  compared  with  8  dollars,  and  5 
feet  can  be  compared  with  8  feet.  The  speed  of  one  shaft  can  be 
compared  with  the  speed  of  another  shaft. 

147.  If  both  terms  of  a  ratio  be  multiplied  or  both  be  divided 
by  the  same  number,  it  will  not  alter  the  value  of  the  ratio. 
Thus,  the  value  of  the  ratio  32  :  12  is  2§;  multiplying  both  terms 
by  3,  the  ratio  becomes  96  :  36,  and  its  value  is  2§,  as  before; 
dividing  both  terms  by  4,  the  ratio  becomes  8  :  3,  and  its  value 
is  2f ,  as  before.     The  reason  for  this  is  seen  when  the  ratio  is 

32 
written  in  the  fractional   form,   y~.     Now    regarding   this   ex- 
pression as  a  fraction,  it  has  been  shown  in  connection  with  frac- 
tions that  multiplying  or  dividing  the  numerator  and  denominator 
by  the  same  number  does  not  alter  the  value  of  the  fraction. 


PROPORTION 

148.  If  there  are  two  ratios,  each  having  the  same  value,  and 
they  are  written  so  as  to  indicate  that  the  ratios  are  equal,  the 
resulting  expression  is  called  a  proportion.     Thus,  the  value  of  the 

ratio  Q-r is  f,  and  the  value  of  the  ratio  ~  '7fi  is  f ;  placing 

5  hours       $1.10 
these  two  ratios  equal  to  each  other,  .,  ,  —  ■  -  =  .  an  ex- 

pression that  is  called  a  proportion. 

It  is  to  be  noted  that  the  value  of  any  ratio  is  always  an  atari  raci 
number  (Art.  6) ;  it  is  for  this  reason  that  the  ratio  of  two  concrete 
numbers  can  be  placed  equal  to  the  ratio  of  two  entirely  different 
concrete  numbers,  as  in  the  above  case.  Hence,  when  stating  a 
proportion,  it  is  not  customary  to  write  the  name  of  the  quanti- 
ties, and  the  above  proportion  would  ordinarily  be  written  either 

as  'e  =  rrc  or  as  5  : 8  =  1.10  :  1.76.     Here  5  is  called  the  first 
o        1 .7b 


80  ARITHMETIC  §1 

term,  8  is  called  the  second  term,  1.10  is  called  the  third  term, 
and  1.76  is  called  the  fourth  term.  When  written  in  the  second 
form  above,  the  two  outside  terms,  5  and  1.76,  are  called  the 
extremes,  and  the  two  inside  terms,  8  and  1.10,  are  called  the 
means. 

149.  Law  of  Proportion. — In  any  proportion,  the  product  of  the 
extremes  is  equal  to  the  product  of  the  means;  thus,  in  the  propor- 
tion just  given,  5  X  1.76  =  8.8,  and  8  X  1.10  =  8.8. 

To  understand  the  reason  for  this  law,  the  value  of  the  two 
ratios  in  the  above  proportion  is  .625,  and  the  proportion  may  be 
reduced  to  the  expression  .625  =  .625.  Dividing  both  numbers 
by  .625,  the  expression  reduces  to  1  =  1.  Now  writing  the  pro- 
portion in  the  form  ^  =  t^,  divide  both  ratios  by  %,  and  the 

,    .    5       8        1.10       8  8.8 

result  is  8  X  ^  =  ^Tq  X  5'  or  !   =  88  =  L  '       X 

is  the  product  of  the  means  and  5  X  1.76  is  the  product  of  the 

extremes,  and  both  are  equal  to  8.8. 

There  is  still  another  way  of  writing  a  proportion  that  was 

formerly  used  by  the  older  writers  on  mathematical  subjects; 

they  employed  the  double  colon  in  place  of  the  sign  of  equality 

when     writing     the    proportion    in    the    second    form;    they 

15       405 
would  express  the  proportion  y~  =  qo7  as  15  :  12  ::405  :  324. 

Here,  as  before,  the  product  of  the  extremes  equals  the  product 
of  the  means,  since  15  X  324  =  12  X  405  =  4860.  This  last 
way  of  writing  a  proportion  is  not  much  used  at  this  time,  the 
sign  of  equality  being  preferred. 

150.  A  proportion  may  be  read  in  two  waj^s.  Consider  the 
proportion  16  :  10  =  88  :  55;  this  may  be  read  16  is  to  10  as  88 
is  to  55,  or  it  may  be  read  the  ratio  of  16  to  10  equals  the  ratio  of 
88  to  55.  Either  way  is  correct,  but  the  second  is  to  be  preferred 
when  using  the  fractional  form  for  expressing  the  ratios. 

151.  A  proportion,  like  a  ratio,  may  be  direct  or  inverse;  in  a 
direct  proportion,  both  ratios  are  direct,  but  in  an  inverse  propor- 
tion, one  of  the  ratios  is  inverse.  If  both  ratios  were  inverse,  the 
proportion  would  become  direct  again.  For  example,  the  state- 
ment that  the  ratio  of  16  to  10  equals  the  inverse  ratio  of  55  to  88 
indicates  an  inverse  proportion.  To  write  it,  first  state  it  as 
though  it  were  a  direct  proportion  and  then  invert  one  of  the 


§1  RATIO  AND  PROPORTION  81 

ratios;  thus,  16  :  10  =  55  :  88.  Now  inverting  the  first  ratio, 
10  :  16  =  55  :  88,  whence,  10  X  88  =  16  X  55;  or,  inverting  the 
second  ratio,  16  :  10  =  88  :  55,  whence,  16  X  55  =  10  X  88. 
The  proportion  as  first  written  was  not  true,  since  16  X  88 
=  1408  and  10  X  50  =  550;  but  by  inverting  one  of  the  ratios,  it 
became  true.  The  test  of  any  proportion  is  the  law  of  Art.  149. 
//  the  product  of  the  extremes  does  not  equal  the  product  of  the  mearis, 
then  the  expression  is  not  a  proportion. 

152.  To  Find  the  Value  of  One  Unknown  Term  in  a  Proportion. 
The  object  of  any  proportion  is  to  find  the  value  of  one  of  the 
terms  when  the  values  of  the  other  three  are  known;  this  fact  will 
be  made  clearer  shortly. 

Let  x  represent  the  value  of  the  term  that  is  not  known,  and 
suppose  the  proportion  is  14  :  8  =  49  :  x.  By  the  law  of  pro- 
portion, 14  X  x  =  8  X  49.     Now,  evidently,  if  14  times  x  =  8 

8  X  49 
X  49,  1  times  x  must  equal  — rj —  =  28,  and  the  proportion 

14  :  8  =  49  :  28    is    true,    because    14  X  28  =  8  X  49  =  392. 

Suppose  the  first  term  had  been  unknown;  then  x  :  8  =  49  :  28, 

8  X  49 
and  1  times  x  =  x  =       „„       =14.     Here  it  is  seen  that  if  the 

unknown  is  one  of  the  extremes,  its  value  can  be  found  by  dividing 
the  product  of  the  means  by  the  other  extreme. 

Suppose  the  second  term  had  been  unknown;  then  the  propor- 
tion  would    have    been   14  :  x  =  49  :  28,  from  which   14  X  28 

14  X  28 
=  49  X  re,  and  the  value  of  x  is  evidently  -    .  (.  - -  =  x,  or  x  =  8. 

Lastly,  suppose  that  the  third  term  had  been  unknown;  then  14: 

14  V  28 
8  =  x  :  28,  and  the  value  of  x  is  evidently ~ =  x,  or  x 

=  49.  Here  it  is  seen  that  if  the  unknown  is  one  of  the  means,  its 
value  can  be  found  by  dividing  the  product  of  the  extremes  by  the 
known  mean. 

153.  Proportion  is  used  for  solving  a  great  variety  of  problems. 
For  example,  the  statement  that  "the  circumferences  of  any  two 
circles  are  to  each  other  as  their  diameters"  or  that  "the  circum- 
feience  of  a  circle  varies  directly  as  its  diameter"  implies  a  propor- 
tion. In  the  first  statement,  if  the  diameter  and  circumference 
of  any  one  circle  are  known  and  the  diameter  (or  circumference) 
of  some  other  circle  is  given,  the  circumference  (or  diameter)  of 
that  circle  may  be  found  by  the  method  explained  in  Art.  152. 


82  ARITHMETIC  §1 

Thus,  the  circumference  of  a  circle  whose  diameter  is  5  inches  is 
15.708  inches,  very  nearly;  hence,  to  find  the  circumference  of 
a  circle  whose  diameter  is  17.6  inches,  first  form  the  proportion 

5  :  17.6  =  15.708  :  x,  and  x  =  ,     ' =  55.29216  inches. 

Note  particularly  the  way  in  which  the  above  proportion  was 
formed.  The  first  ratio  consists  of  the  two  diameters,  which  are 
like  quantities,  and  the  second  ratio  consists  of  the  two  circum- 
ferences, which  are  also  like  quantities.  The  first  and  third 
terms  in  every  direct  proportion  must  belong  or  relate  to  the  same 
thing,  and  the  second  and  fourth  terms  must  also  belong  or  relate 
to  the  same  thing;  in  the  present  case,  5  and  15.708  are  the  diame- 
ter and  circumference  of  one  circle,  and  17.6  and  55.29216  are  the 
diameter  and  circumference  of  the  other  circle.  It  will  also  be 
noted  that  the  second  circle  is  larger  than  the  first  circle ;  hence, 
its  circumference  must  also  be  larger.  But,  if  #  be  written  for 
the  third  term  of  the  proportion,  the  value  that  will  be  found  for 
x  will  be  smaller  than  the  third  term,  thus  showing  again  that  x 
must  be  written  as  the  fourth  term.  If  desired,  the  proportion 
might  have  been  written  15.708  :  x  =  5  :  17.6,  and  the  same 
value  will  be  obtained  for  x.  It  is  customary,  however,  to  have 
the  first  ratio  consist  of  numbers  or  quantities  that  are  known. 

Now  referring  to  the  second  statement  above,  that  the  circum- 
ference of  a  circle  varies  directly  as  its  diameter,  this  means  that 
if  the  diameter  increases,  the  circumference  also  increases,  and 
if  the  diameter  decreases,  the  circumference  also  decreases,  both 
increasing  or  decreasing  in  the  same  ratio.  Therefore,  if  the 
circumference  of  a  circle  whose  diameter  is  5  inches  is  15.708  inches, 
and  it  is  desired  to  find  the  circumference  of  the  circle  when  the 
diameter  is  increased  to  17.6  inches,  form  the  proportion  5  :  17.6 
=  15.708  :  x.  Substituting  the  value  of  x  in  this  proportion, 
5  :  17.6  =  15.708  :  55.29216.  The  diameter  of  the  second  circle 
has  increased  in  the  ratio  17.6  :  5,  whose  value  is  3.52,  and  the 
circumference  of  the  second  circle  has  increased  in  the  ratio 
55.29216  :  15.708,  whose  value  is  3.52;  hence,  the  diameter  and 
the  circumference  increased  in  the  same  ratio.  From  the  fore- 
going, it  is  seen  that  the  word  vary,  as  here  used,  always  implies  a 
proportion  or  that  a  proportion  can  be  formed. 

In  practice,  instances  frequently  occur  in  which  an  increase  in 
one  of  the  terms  of  the  first  ratio  results  in  a  decrease  in  the  value 
of  the  corresponding  term  in  the  second  ratio,  and  vice  versa. 


§1  RATIO  AND  PROPORTION  83 

Thus,  suppose  a  certain  volume  of  air  be  confined  in  a  cylinder, 
and  that  it  is  made  to  keep  this  volume  by  means  of  a  piston  that 
is  free  to  move  up  and  down  and  on  which  is  placed  a  weight. 
Now  it  is  evident  that  if  the  weight  be  increased,  the  piston  will 
move  downward  and  the  volume  of  the  air  will  be  less;  or,  if  the 
weight  be  decreased,  the  piston  will  move  upward  and  the  volume 
of  the  air  will  be  greater.     In  other  words,  the  volume  decreases 
as  the  pressure  increases,  and  vice  versa.     This  fact  is  expressed 
more  elegantly  by  saying  that  the  volume  varies  inversely  as  the 
pressure,  which,  of  course,  implies  an  inverse  proportion,  when  a 
proportion  is  possible.     As  an  example,  it  is  known  that  when  the 
temperature  remains  the  same,  the  volume  of  a  gas    (air,  for 
instance)  varies  inversely  as  the  pressure;  if  the  volume  is  12.6 
cubic  feet  when  the  pressure  is  14.7  pounds  per  square  inch,  what 
is  the  volume  when  the  pressure  is  45  pounds  per  square  inch? 
Forming  the  proportion  as  though  it  were  direct,  14.7  :  45  =  12.6 
:  x.     Here  it  is  seen  that  the  value  of  x  obtained  from  the  pro- 
portion will  be  greater  than  12.6,  and  it    should  be  less,  thus 
indicating  an  inverse  proportion.     Since  the  proportion  is  inverse, 
invert  the  terms  of  one  of  the  ratios,  say  the  second,  obtaining 

14.7  X  12.6       .  ...      .. 

14.7  :45  =  x  :  12.6,  from   which  x  =  ^ =  4. 116  cubic 

feet.     Note  that  the  volume  is  smaller  than  the  original  volume, 
as  it  ought  to  be. 

Example  1. — If  5  men  can  do  a  certain  piece  of  work  in  22  hours,  how  long 
will  it  take  9  men  to  do  the  same  work,  if  they  all  work  at  the  same  rate? 

Solution. — It  is  evident  that  9  men  can  do  the  work  in  less  time  than 
5  men;  hence,  the  proportion  is  inverse.  Stating  as  a  direct  proportion,  5  : 
9  =  22:  x;  then,  inverting  the  second    ratio,   5  :  9  =  x  :  22,  from  which 

e  Y  22 
x  =        g        =  12%  hours.     Ans. 

Example  2.— If  5  men  earn  $48.95  in  22  hours,  how  much  will  9  men  earn 
in  the  same  time  at  the  same  rate  of  pay? 

Solution.— It  is  here  plain  that  9  men  will  receive  a  greater  wage  than  5 
men;   hence,   the  proportion  is  direct,  and  5:9=  48.95  :  x,  from  which 

=  48.95  X  9  _  n      An^     Note  tha(.  the  timGj  .22  hours,  has  nothing 

to  do  with  the  proportion,  because  it  is  the  same  in  both  cases. 

154.  Compound  Proportion. — A  simple  proportion  is  one  in 
which  all  the  terms  consist  of  but  one  number  in  each  term ;  but 
when  two  of  the  terms  or  all  four  of  .the  terms  contain  more  than 
one  number,  then  the  proportion  is  said  to  be  compounded,  or  it  is 
called  a  compound  proportion.     Suppose  the  last  example  had 


84  ARITHMETIC  §1 

been  stated  thus:  if  5  men  earn  $48.95  in  22  hours,  how  much 
will  9  men  earn  in  15  hours?  Here  5  men  in  22  hours  earn  $48.95, 
and  it  is  required  to  find  how  much  9  men  will  earn  in  15  hours, 
all  being  paid  at  the  same  rate.  The  proportion  in  this  case 
should  be  stated  as  follows:  5  X  22  : 9  X  15  =  48.95  :  x,  from 

which  x  =  ^  y  99    ' —  =  $60,073^.     The  reason  for  stating 

the  proportion  in  this  manner  is  that  5  men  working  22  hours  is 
the  same  as  one  man  working  5  X  22  =  110  hours,  and  during 
this  110  hours,  $48.95  was  earned.  Also,  9  men  working  15 
hours  is  the  same  as  one  man  working  9  X  15  =  135  hours,  and 
during  this  135  hours,  a  certain  amount  was  earned  that  is  re- 
presented by  x.  Instead  of  multiplying  before  substituting  in 
the  proportion,  it  is  better  to  indicate  the  multiplication,  so  as 
to  take  advantage  of  any  opportunity  of  cancelation;  thus, 

3        4  45 

-  -  <^mF  -  — • 

2 

Example. — If  25  men  can  dig  a  ditch  600  feet  long,  \y2  fept  wide,  and  3  V£ 
deep  in  a  certain  number  of  dajrs,  working  S  hours  a  day,  how  long  a  ditch 
4  feet  wide  and  4  feet  deep  can  36  men  dig  when  working  10  hours  a  day? 

Solution. — If  men  and  hours  be  considered  in  forming  the  first  ratio, 
the  first  term  is  compounded  of  25  men  and  S  hours,  representing  time;  the 
third  term  is  compounded  of  600  feet,  4.5  feet,  and  3.5  feet,  representing 
work  done;  the  second  term  is  compounded  of  36  men  and  10  hours;  and  the 
fourth  term  is  compounded  of  a;  feet,  4  feet,  and  4  feet.  Then,  25  X  8  :  36 
X  10  =  600  X  4.5  X  3.5  :  x  X  4  X  4,  from  which 

36  X  10  X  600  X4.5  X  3.5  , 

X  = 26X8XTX4 =  106SH  feet"     AnS- 


EXAMPLES 

Find  the  value  of  x  in  the  following  proportions : 

(1)  6  :  8  =  x  :  18.  Ans.  13.5. 

(2)  20  :  x  =  8  :  26.  Ans.  65. 

(3)  x  :  7  =  81  :  91.  Ans.  6K3. 

(4)  11  :  16  =  517  :  x.  Ans.  752. 

(5)  5.5  X  18  :  7.2  X  15  =  4.4  X  25  :  10.5  X  j?  X  x.  Ans.  7&. 

(6)  120  X  64  :  540  X  328  =  110  :  x.  Ans.  2536%. 

(7)  If  a  certain  supply  of  provisions  will  last  525  men  129  days,  how  long 
will  it  last  603  men?  Ans.  112  +  days. 


§1  RATIO  AMD  PROPORTION  85 

(8)  Referring  to  Art.  153,  suppose  the  volume  of  an  air  compressor  is 
1.95  cubic  feet.  If  the  cylinder  be  filled  with  air  at  a  pressure  of  14.7 
pounds  per  square  inch,  and  the  air  is  compressed  until  the  volume  is  .31 
cubic  feet,  what  is  the  pressure,  the  temperature  remaining  the  same? 

Ans.  92.47—  pounds  per  square  inch. 

(9)  Referring  to  Art.  163*  what  is  the  circumference  of  a  circle  whose 
diameter  is  12%  inches?  Ans.  38.8773  inches. 


ARITHMETIC 

(PART  2) 

EXAMINATION  QUESTIONS 

(1)  What  is  the  greatest  common  divisor  of  15,862  and  1300? 

Ans.  77. 

(2)  Find  the  G.C.D.  of  45, 135,  270  and  405.  Ans.  45. 

(3)  Find  the  L.C.M.  of  45,  135,  270  and  405.  Ans.  810. 

(4)  Which  fraction  is  the  larger,  ^  or  s\\?  Ans.  ^jV 

(5)  What  is  the  value  of  f  +  &  +  U  +  ft  -  fj? 

A?is.  liVi. 

(6)  Reduce  the  following  fractions  to  their  lowest  terms:  (a) 
fif;  (&)  tWV;  (c)  \il;  (d)  what  is  the  difference  between  the  frac- 
tions (a)  and  (b)  ?  (e)  what  is  the  sum  of  the  fractions  (6)  and  (c)  ? 

Ans.  (a)  f;  (6)  *?;  (c)  &;  (d)  ^T;  (e)  |HL 

(7)  What  is  the  value  of  \\  -  ft  +  ft  -  H  +  ixr? 

(8)  What  is  the  product  of  the  sum  and  the  difference  of  4| 
and3£?  Ans.  13,VuV. 

(9)  A  carload  of  coal  weighing  47,960  pounds  was  delivered  to 
a  paper  mill;  T\ths  was  used  the  first  week,  Aths  the  second  week, 
and  T3-jths  the  third  week;  how  many  pounds  then  remained? 

Ans.  8502  pound. 

(10)  What  must  be  paid  for  8|o  ounces  of  dye  if  the  cost  is 
$1.27  per  ounce?  Ans.  $11.19. 

(11)  Divide  (a)  0.67  by  0.0007042;  (6)  18|  by  653.109. 

A         f  (a)  951.43+. 
Am-    1(6)   .028135-. 

(12)  The  distance  across  the  diagonally  opposite  corners  of  a 
hexagonal  (six-sided)  nut  is  called  the  outside'  diameter,  and  for 
an  unfinished  nut  is  found  as  follows:  to  L|  times  the  diameter  of 
the  bolt  add  |  inch,  and  multiply  the  sum  by  1.1547.  What  is 
the  outside  diameter  of  a  nut  for  a  1-J-  inch  bolt?  Give  result  to 
nearest  64th  of  an  inch.  Ans.  2e"V  =  2^  inches. 

87 


88  ARITHMETIC  §1 

147  X  n  -1-  X  -2- ' fi  X    s  8 

(13)  Simplify  the  complex  fraction,  — w^va  Z    2a'v    5  2US- 

ouo-i  a  xtt  <*  T~2J> 

Ans.  65V 

(14)  The  area  of  a  circle  is  .7854  timet  the  square  of  the  di- 
ameter; what  is  the  area  of  a  circle  whose  diameter  is  5T\? 

Ans.  21.135+. 

(15)  What  is  the  value  of  6{9  -  [14  -  4(12  -  7) (11  +  5) 
-h  48]  +  7(45  -  51  -  13)}  ?  Ans.  304. 

(16)  Find  the  value  of  x  in  the  proportions:  (a)  15  :  21  = 
48  :  x;  (b)  18  : 8  =  x  :  54;  (c)  12  :  x  =  128  :  96. 

(  (a)  x  =    67.2. 
Ans.       (b)  x  =--  121.5. 
[(c)   x  =    72. 

(17)  If  5  men  can  do  a  certain  piece  of  work  in  8  days,  how 
long  will  it  take  10  men  to  do  3  times  as  much  work? 

Ans.  12  days. 

(18)  If  the  pressure  of  steam  in  an  engine  cylinder  varies  in- 
versely as  the  volume,  and  the  pressure  is  110  pounds  per  square 
inch  when  the  volume  is  0.546  cubic  feet,  what  is  the  pressure 
when  the  volume  is  2.017  cubic  feet? 

Ans.  29.78—  pounds  per  square  inch. 


ARITHMETIC 

(PART  3) 


SQUARE  ROOT 

166.  The  power  of  a  number  has  been  previously  defined;  it  is 
the  continued  product  of  as  many  equal  factors  as  are  indicated 
by  the  exponent;  a  very  important  problem  is:  given  the  power,  to 
find  what  equal  factor  was  used  to  produce  the  power.  The 
number  so  found  is  called  the  root  of  the  given  number.  The 
roots  arc  named  in  the  same  manner  as  the  powers.  For  example, 
if  two  equal  numbers  are  used  to  form  the  power,  one  of  them  is 
called  the  square  root  of  the  given  number;  if  three  equal  factors 
are  used  to  form  the  power,  one  of  them  is  called  the  cube  root;  if 
four  equal  factors  are  used  to  form  the  power,  one  of  them  is  called 
the  fourth  root;  etc.  Thus,  the  square  root  of  9  is  3,  because 
3X3  =  9;  the  cube  root  of  125  is  5,  because  5X5X5  =  125;  etc. 

156.  The  root  of  a  number  is  indicated  by  writing  before  it 
what  is  called  the  radical  sign  y/ ,  and  to  specify  what  root,  a 
small  figure  is  written  in  the  opening  of  the  sign ;  thus,  \/,  v7,  \/ , 
etc.  indicate  the  square  root,  cube  root,  and  fifth  root,  respec- 
tively. The  square  root  is  indicated  so  frequently  that  it  is 
customary  to  omit  the  small  figure,  which  is  called  the  index  of  the 
root,  using  only  the  sign;  hence,  y/9,  \/81,  \/l44,  etc.  indicate 
respectively  the  square  root  of  9,  the  square  root  of  81,  the  square 
root  of  144,  etc.  The  index  cannot  be  omitted  when  indicating 
any  other  root  than  the  square  root.  It  is  also  customary  to  use 
the  vinculum  in  connection  with  the  radical  sign.  Thus,  -^1728, 
indicates  the  cube  root  of  1728;  v/777(l  indicates  the  fifth  root 
of  7776;  v/24  X  54  indicates  the  square  root  of  the  product  of  24 
and  54.  If  it  were  desired  to  indicate  the  product  of  54  and  the 
square  root  of  24,  it  may  be  done  in  two  ways,  either  as  54\/24 
or  as  v/24  X  54.  It  will  be  observed  that  in  the  first  form, 
the  sign  of  multiplication  is  omitted  between  54  and  the  radical 
sign;  this  is  customary,  and  when  so  written,  multiplication  is 
always  understood. 

89 


90  ARITHMETIC  §1 

The  root  of  a  number  is  also  frequently  indicated  by  using  a 
fractional  exponent,  the  denominator  of  the  fraction  indicating 
the  root.  For  example,  196'  has  the  same  meaning  as  V196, 
243 l  =  ^243,  343  *  =  s/343,  etc.  If  the  numerator  of  the  frac- 
tional exponent  is  some  number  other  than  1,  it  indicates  that  the 
number  is  to  be  raised  to  the  power  indicated  by  the  numerator 
and  the  root  taken  that  is  indicated  by  the  denominator.  Thus, 
243'  indicates  that  243  is  to  be  raised  to  the  fourth  power  and  the 
fifth  root  is  then  to  be  found,  or  it  means  that  the  fifth  root  of  243 
is  to  be  found  and  the  result  raised  to  the  fourth  power;  the  final 
result  will  be  the  same.  For  instance,  v/243  =  3,  and  34  =  81; 
also,  3434  =  3,486,784,401  and  -y/3,486,784,401  =  81,  because 
815  =  3,486,784,401.  In  other  words,  243*  =  (-^243)4or  x/^S*, 
and  the  value  of  both  these  expressions  is  81.  Both  expressions 
may  also  be  written  (243 J)4  or  (2434)\ 

157.  A  number  is  said  to  be  a  perfect  square,  a  perfect  cube,  a 
perfect  fifth  power,  etc.,  when  its  square  root,  cube  root,  fifth  root, 
etc.,  can  be  expressed  exactly;  for  instance,  196  is  a  perfect  square, 
because  \/l96  =  14;  343  is  a  perfect  cube,  because  \/343  =  7; 
3,486,784,401  is  a  perfect  fifth  power,  because  ^3,486,784,401 
=  81.  Only  comparatively  few  numbers  are  perfect  powers ;  thus, 
between  1  and  100,  the  only  perfect  powers  are  1,  4,  8,  9,  16,  25, 
27,  32,  36,  49,  64,  81,  and  100,  a  total  of  13  only.  Numbers 
that  are  not  perfect  powers  are  called  imperfect  powers;  thus 
7,  12,  26,  47,  etc.  are  imperfect  powers.  It  is  to  be  noted  that  a 
number  may  be  perfect  for  some  particular  power,  but  not  for 
others;  in  fact,  this  is  usually  the  case.  For  example,  243  is  a 
perfect  fifth  power,  but  it  is  not  perfect  for  any  other  power;  16 
=  24  and  42,  and  is  a  perfect  fourth  power  and  a  perfect  square; 
64  =  26  =  43  =  82,  and  is  a  perfect  sixth  power,  a  perfect  cube, 
and  a  perfect  square.  Numbers  that  are  perfect  for  more  than 
one  power  are  rare. 

The  root  of  an  imperfect  power  is  never  exact;  the  root  of  such 
a  number  may  be  calculated  to  any  number  of  figures,  in  the  same 
way  that  the  quotient  may  be  carried  to  any  number  of  decimal 
places  in  division  when  the  divisor  contains  some  factor  other 
than  2  or  5. 

158.  The  square  root  of  a  number  is  very  frequently  desired ; 
occasionally,  some  other  root  is  required,  but  the  process  of  find- 
ing any  root  other  than  the  square  is  so  laborious,  and  such  roots 


§1  SQUARE  ROOT  91 

are  required  so  seldom,  that  only  the  method  for  finding  the 
square  root  of  any  number  is  described  here.  When  other  roots 
are  desired,  they  are  usually  found  by  using  tables  of  logarithms. 
A  method  for  finding  cube  and  fifth  roots  is  described  in  Elemen- 
tary Applied  Mathe?natics. 

159.  There  are  many  methods  of  finding  the  square  root  of  a 
number,  the  best  general  method  being  that  known  as  Horner's 
method,  which  is  the  one  that  will  be  explained  here. 

The  first  step,  no  matter  what  method  is  used,  is  to  point  off  the 
number  whose  root  is  to  be  found  into  periods  of  two  figures  each, 
beginning  at  the  decimal  point  and  going  to  the  left,  and  to  the 
right  also,  in  the  case  of  mixed  numbers  and  pure  decimals.  For 
this  purpose,  it  is  best  to  use  a  tick— a  mark  something  like  an 
apostrophe.  For  example,  the  number  11,778,624  is  pointed  off 
as  11'77'86'24;  the  numbers  .7854  and  .0003491  would  be  pointed 
off  as  .78'54  and  .00'03'49'10;  the  number  755.29216  would  be 
pointed  off  as  7'55.29'21'60.  When  the  right  hand  period  of  the 
decimal  is  not  complete,  that  is,  does  not  contain  two  figures,  add 
a  cipher.  The  left-hand  period  of  the  integral  part  may  contain 
either  one  or  two  figures,  according  to  whether  the  number  of 
figures  in  the  integral  part  is  even  or  odd. 

Suppose   the   square   root   of    11,778,624   is   required.     First 

point  off  the  number  into  periods  of 
two    figures    each.     The    work    is 
3  11'77'S6'24(3432  done  in  two  columns,   as  follows: 

3^       _9  The  first  period,  11,  of  the  given 

60         277  number  is  not  a  perfect  square,  and 

_i         2^L  the  largest  perfect  square  less  than 

64  2186  1 1  is  9,  the  square  root  of  which  is 

-i  20i?  3.     Write  3  as  the  first  figure  of  the 

68*J  1S724  r00t'  in  tllG  SamG  mann0r  aS  Wnen 

—  finding   the   quotient   in   division, 

3  and  also  write  it  at  the  head  of  the 

^0  first   column.     Now   multiply   the 

2  3  in  the  first  column  by  the  3  in  the 

6862  root,  and  write  the  product  under 

the  first  period;  subtract,  obtaining 

a  remainder  of  2.     Add  the  first 

figure  of  the  root  to  the  number  in  the  first  column,  and  the  sum  is 

6,  to  which  annex  a  cipher,  making  it  60;  also  annex  the  second 

period  of  the  given  number  to  the  remainder,  2,  in  the  second 


92  ARITHMETIC  §1 

column,  obtaining  as  a  result  277.  Divide  the  last  number  in  the 
second  column,  277,  by  the  last  number  in  the  first  column,  60, 
and  the  quotient  is  4,  which  is  probably  the  next  figure  of  the  root. 
Write  4  for  the  second  figure  of  the  root,  add  it  to  60,  the  last 
number  in  the  first  column,  obtaining  64;  multiply  64  by  4,  the 
second  figure  of  the  root,  and  subtract  the  product,  256,  from  the 
last  number  in  the  second  column,  277,  obtaining  21  for  the 
remainder.  Then  add  the  second  figure  of  the  root,  4,  to  64,  the 
last  number  in  the  first  column,  and  the  sum  is  68,  to  which  annex 
a  cipher,  making  it  680.  Also  annex  the  third  period,  86,  to  the 
remainder  21,  in  the  second  column,  making  it  2186.  Divide 
2186  by  680,  and  the  quotient,  3,  is  probably  the  next,  or  third, 
figure  of  the  root.  Write  3  for  the  third  figure  of  the  root,  add  it 
to  680  in  the  first  column,  obtaining  683,  and  multiply  683  by  the 
third  figure  of  the  root;  the  product,  2049,  is  subtracted  from  2186, 
and  the  remainder  is  137.  Add  the  third  figure  of  the  root,  3,  to 
683,  obtaining  686,  to  which  annex  a  cipher,  making  it  6860. 
Annex  the  fourth  (and  last)  period  of  the  given  power  to  the  last 
remainder,  making  it  13724;  divide  this  by  6860,  and  the  quotient 
2,  is  the  fourth  figure  of  the  root.  Add  the  fourth  figure  of  the 
root  to  6860,  obtaining  6862;  multiplying  this  by  the  fourth  figure 
of  the  root,  the  product  is  13724,  which  subtracted  from  the  last 
number  in  the  second  column  gives  a  remainder  of  0.  Therefore, 
VTl,778.624  =  3432.     Ans. 

Note  that  the  first  column  is  formed  entirely  by  adding,  and 
that  two  additions  are  made  for  each  figure  of  the  root.  The 
above  explanation  of  the  method  will  be  made  much  clearer  to 
the  reader  if  he  will  take  pencil  and  paper  and  set  down  the  vari- 
ous numbers  exactly  as  they  occur  in  the  explanation.  After  a 
few  examples  have  been  solved,  the  method  will  be  clear,  and  it 
will  be  found  easy  to  remember;  in  fact,  the  work  will  be  found  to 
be  but  little  harder  than  ordinary  division. 

160.  Referring  to  the  last  example,  the  numbers  in  the  first 
column  (the  ones  to  which  the  ciphers  have  been  annexed)  that 
are  used  as  divisors  to  determine  figures  of  the  root  are  called 
trial  divisors;  the  numbers  that  they  divide  are  called  trial 
dividends,  and  the  numbers  that  are  obtained  by  multiplying 
the  trial  divisors  by  the  root  figures  may  be  called,  for  the  want 
of  a  better  term,  the  partial  squares  or  partial  powers.  When 
the  given  number  is  a  perfect  square,  the  sum  of  the  partial 
powers,  added  as  they  stand,  must  equal  the  given  power;  thus, 


§1  SQUARE  ROOT  93 

the  sum  of  9,  250,  2049,  and  13724,  the  partial  powers,  added 

g  as  they  stand,  is  11,778,024,  the  given  number. 

256  In  this  example,  the  first  trial  divisor  is  00, 

2049  aiid  the  first  trial  dividend  is  277;  the  second 

^  trial  divisor  is  080,  and  the  second  trial  divi- 

11778624  dend  ig  2186;  etc.     The  numbers  obtained  by 

adding  the  root  figures  to  the  trial  divisors  are  called  the  complete 
divisors;  64,  083,  and  0802  are  the  complete  divisors. 

When  finding  the  second  figure  of  the  root,  it  may  happen  that 
the  product  of  the  complete  divisor  and  the  second  figure  of  the 
root  is  greater  than  the  partial  power;  in  such  case,  try  a  number 
one  unit  smaller  for  the  second  figure  of  the  root.  If  the  product 
of  the  new  complete  divisor  and  the  new  second  figure  of  the  root 
is  still  greater  than  the  trial  divisor,  reduce  the  second  figure  of 
the  root  one  unit  more.  It  will  never  be  necessary  to  make  more 
than  two  trials,  and  seldom  more  than  one. 

Example. — Find  the  square  root  of  12,054,784. 
3  12'05'47'84  (3472.  Ans.        Solution.— The  work  is  almost  exactly 

3  9  the  same  as  in  the  preceding  example.     It 

g0  305  is  to  be  noted,  however,  that  the  quotient 

obtained  by  dividing  the  first  trial  dividend 
by  the  first  trial  divisor  is  5;  but  when  5  is 
added  to  60  for  the  complete  divisor,  and 
this  is  multiplied  by  5,  the  product  is 
greater  than  the  trial  dividend;  hence,  4 
is  tried  for  the  second  figure  of  the  root. 


161.  If  the  given  number  whose  root  is  to  be  found  contains  a 
decimal  there  will  be  as  many  integral  places  in  the  root  as  there 
are  periods  in  the  integral  part  of  the  given  number.  Thus,  the 
square  root  of  1205.4784  contains  two  integral  places;  the  square 
root  of  43200.231  contains  three  integral  places,  because  when 
pointed  off,  it  becomes  4'32'0G.23'10,  and  there  are  three  periods 
in  the  integral  part;  etc. 

If  the  given  number  is  a  pure  decimal,  the  root  will  also  be  a 
pure  decimal;  and  if  there  arc  ciphers  between  the  decimal  point 
and  the  first  digit,  there  will  be  as  many  ciphers  between  the 
decimal  point  and  the  first  digit  of  the  root  as  there  are  periods 
containing  no  digits.  Thus,  the  square  root  of  .000081  is  .009, 
because  when  pointed  off,  the  given  number  becomes  .00'00'81, 


4 

256 

64 

4947 

4 

4809 

680 

13884 

7 

13S84 

687 

7 

6942 

'.II 


ARITHMETIC 


2 

40 
_9 
49 
_9 

5800 

(i 

5806 


1 
1 

20 
_8 
28 
_8 
360 
_8 
368 
8 
3760 
6 
3766 
6 
37720 
7 
37727 
7 

377340 


s'll    IS '30  (29.06.  Ans. 

4 

141 

441 

34836 

34836 


and  there  are  two  periods  containing  no  digits;  the  square  root  of 
.000 100  is  .014,  because  when  pointed  off,  the  given  number 
becomes  .00'01'96,  and  there  is  one  period  containing  no  digit; 
etc.  To  prove  this  statement,  square  the  roots;  thus,  .0092 
=  .000081,  and  .0142  =  .000196. 

Example  1. — What  is  the  square  root  of  844.4836? 

Solution. — On  dividing  the  first  trial  dividend,  444,  by  the  first  trial 
divisor,  40,  the  quotient  is  11;  but,  the  second  figure  of  the  root  cannot  be 

greater  than  9;  hence,  try  9  for  the  second 
figure.  The  second  trial  divisor  is  580,  and 
it  is  larger  than  the  second  trial  dividend, 
which  is  348;  consequently,  annex  another 
cipher  to  the  trial  divisor,  making  it 
5800,  and  annex  another  period  to  the 
trial  dividend,  making  it  34836,  which 
contains  the  trial  divisor  6  times.  As 
there  are  two  periods  in  the  integral  part 
of  the  given  number,  there  are  two  figures 
in  the  integral  part  of  the  root. 
Example  2. — Find  the  square  root  of  356  to  three  decimal  places. 

Solution. — The  first  figure  of  the  root  is  1,  since 
the  first  period  is  less  than  4.  When  dividing 
the  first  trial  dividend  by  the  first  trial  divisor, 
the  quotient  is  256  H-  20  =  12  +  ;  but  the 
second  figure  of  the  root  cannot  exceed  9. 
Trying  9,  the  complete  divisor  is  29,  and  29 
X  9  =  261,  which  is  greater  than  the  partial 
power,  256;  hence,  try  8  for  the  second  figure 
of  the  root.  The  remainder  is  256,  showing  that 
the  given  number  is  an  imperfect  power.  The 
work  is  now  continued  by  annexing  cipher 
periods  as  shown.  The  sixth  figure  of  the  root, 
the  figure  in  the  fourth  decimal  place,  is  9; 
hence,  the  root  correct  to  three  decimal  places 
is  18.868  —  .  Adding  the  partial  powers,  first 
writing  them  under  the  last,  264089,  the  sum  is 
355.963689,  which  is  the  square  of  18.867;  if 
the  last  remainder,  36311,  disregarding  the 
decimal  point,  be  added  to  this  sum,  the  total 
is  356,  the  given  number,  showing  that  the  wrork 
is  correct. 


3 '56.  (18.8679 

1 

256 

224 

3200 
2944 

~25600 
22596 


300400 
264089 

3631100 
264089 
22596 
2944 
224 
1^ 

355963689 
36311 

351)000000 


162.  If  the  given  number  is  a  common  fraction,  the  root  may 
be  found  by  extracting  the  square  root  of  the  numerator  and 
denominator  separately;  thus,  \/tV  =  t-  But,  unless  both 
numerator  and  denominator  are  perfect  squares,  it  is  better  to 
reduce  the  fraction  to  a  decimal,  and  find  as  many  figures  of  the 


SQUARE  ROOT 


95 


root  as  are  desired;  usually,  4  or  5  figures  of  the  root  of  an  im- 
perfect power  (not  counting  ciphers  between  the  decimal  point 
and  the  first  digit  of  pure  decimals)  are  sufficient  for  practical 
purposes. 

Example. — Find  the  square  root  of  |  correct  to  four  decimal  places. 

Solution. — Reducing  f  to  a  decimal,  it 
j^ns  becomes  .875.  According  to  Art.  161,  the 
root  is  a  pure  decimal  and  the  first  figure 
of  it  is  a  digit.  After  finding  the  first  two 
figures  of  the  root,  there  are  no  more  periods 
to  the  given  number,  and  the  work  is  con- 
tinued by  annexing  cipher  periods  to  the 
different  trial  divisors,  each  period  con- 
taining two  ciphers.  It  is  evident  that 
the  fifth  figure  of  the  root  is  1;  hence,  the 
root  correct  to  4  decimal  places  is  .9354. 
To  prove  that  no  mistake  has  been  made  in 
the  work,  add  the  partial  powers,  obtaining 
.87497316  =  .93542;  adding  the  last  re- 
mainder the  sum  is  .875;  the  given  number, 
which  shows  that  no  mistake  has  been 
made.  Furthermore,  when  expressed  to 
four  figures,  .87497316  =  .8750,  which  is 
also  correct.  It  is  well  to  check  the  work 
in  this  manner,  as  it  prevents  mistakes. 

163.  If  the  reader  has  followed  each  step  of  the  foregoing 
examples  with  pencil  and  paper,  he  should  be  able  to  extract  the 
square  root  of  any  number. 

Rule  I. — Beginning  at  the  decimal  point,  point  off  the  given 
number  into  periods  of  two  figures  each,  including  the  decimal  part, 
if  any. 

II.  Arrange  the  work  in  two  columns,  the  second  column  contain- 
ing the  given  number.  The  first  figure  of  the  root  is  the  square  root 
of  largest  square  that  does  not  exceed  the  first  period. 

III.  Write  the  first  figure  of  the  root  as  the  first  number  in  the 
first  column,  multiply  it  by  the  first  root  figure,  and  subtract  the 
product  from  the  first  period,  and  annex  to  the  remainder  the  second 
period;  this  is  the  first  trial  dividend.  Add  the  first  root  figure  to 
the  number  in  the  first  column,  and  annex  a  cipher,  thus  obtaining 
the  first  trial  dinsor. 

IV.  Divide  the  trial  dividend  by  the  trial  divisor,  and  the  first 
figure  of  the  quotient  (if  less  than  10)  will  probably  be  the  second 
figure  of  the  root;  add  this  figure  to  the  trial  divisor,  obtaining  the 


9 

9 

180 

.87'50  (.93541 
81 
650 

3 

183 

549 
10100 

3 

9325 

1860 

77500 

5 

74816 

1865 

268400 

5 

.81 

18700 

549 

4 

9325 

18704 

74816 

4 

.87497316 

187080 

2684 

. 87500000 

96  ARITHMETIC  §1 

complete  divisor,  which  multiply  by  the  second  root  figure,  and  sub- 
tract the  product  from  the  trial  dividend,  to  which  annex  the  next 
period  to  form  the  next  trial  dividend.  But  if  this  product  is  greater 
than  the  trial  dividend,  try  a  figure  one  unit  less  than  the  quotient 
just  obtained,  adding  it  to  the  trial  divisor,  and  multiplying  as  before. 
If  Vie  product  is  yet  greater  than  the  trial  divisor,  try  a  root  figure  one 
unit  less  than  the  last.  Having  found  a  satisfactory  complete 
dinsor,  add  to  it  the  root  figure  last  found,  and  annex  a  cipher  to 
form  a  new  trial  dinsor.  Divide  the  second  trial  dividend  by  the 
second  trial  divisor,  and  the  integral  part  of  the  quotient  will  be  the 
next  figure  of  the  root. 

V.  Proceed  in  this  manner  until  all  the  periods  of  the  given 
number  have  been  used.  If  the  number  is  an  imperfect  square, 
limit  the  root  to  five  figures  (counting  the  first  digit  as  one),  unless 
more  figures  are  especially  desired,  annexing  cipher  periods  of  two 
figures  each,  if  necessary. 

VI.  7/ at  any  time,  the  trial  divisor  is  larger  than  the  trial  dividend, 
annex  a  cipher  to  the  trial  divisor  and  annex  another  period  to  the 
trial  dividend,  placing  a  cipher  in  the  root.  Proceed  in  this  manner 
until  the  root  has  been  found  or  as  many  figures  have  been  found  as 
are  desired. 

VII.  If  the  quotient  obtained  by  dividing  the  first  trial  dividend 
by  the  first  trial  divisor  is  greater  than  9,  try  9  for  the  second  figure 
of  the  root.  The  sum  of  the  partial  powers  added  as  they  stand, 
must  a  ways  be  equal  to  the  square  of  the  root  as  found,  if  no  mistake 
has  been  made  in  the  work,  and  this  sum  plus  the  last  remainder 
must  equal  the  given  number.  It  is  a  good  plan  to  check  the  work 
in  every  instance  by  ascertaining  if  this  is  the  case. 


EXAMPLES 

(1)  Find  the  square  root  of  293,005.69.  Ans.  541.3 

(2)  Find  VlO  to  four  decimal  places.  Ans.  3.1623- 

(3)  Find  V3.1416  to  four  decimal  places.  Am.  1.7725- 

(4)  Find  V36U  to  four  decimal  places.  Ans.  6.0557  + 

(5)  Find  V.02475  to  five  decimal  places.  Ans.  .15732  + 

(6)  Find  \/|  to  five  decimal  places.  Ans.  .81650  — 

(7)  Find  V. 0000000217  to  five  decimal  places.  Ans.  .00014731  - 


§1  SQUARE  ROOT  97 

APPLICATIONS  OF  SQUARE  ROOT 

164.  In  problems  relating  to  mensuration,  mechanics,  strength 
of  materials,  and  engineering  generally,  it  is  required  very  fre- 
quently to  find  the  square  root  of  numbers.  A  few  applica- 
tions will  be  given  here  by  means  of  examples. 

Example  1. — If  the  area  of  a  circle  is  known  or  given,  the  diameter  of  the 
circle  can  be  found  by  multiplying  the  square  root  of  the  area  by  1.1284. 
What  is  the  diameter  of  a  circle  having  an  area  of  50.28  square  inches? 
Give  result  to  four  decimal  places. 

Solution. — Since  the  result  is  required  to  four  decimal  places,  and  there 
are  four  decimal  places  in  the  number  1.1284  and  one  integral  place,  making 
five  figures  in  all,  find  the  square  root  of  56.28  to  5  +  1  =6  figures,  obtain- 
ing 7.50200-.  Then,  the  diameter  is  1.1284X7.502=8.4652568,  or 
8.4653—  inches  to  four  decimal  places.     Ans. 

Example  2. — When  a  heavy  body  falls  freely,  it  strikes  the  ground  with 
a  velocity  in  feet  per  second  equal  to  8.02  times  the  square  root  of  the  height 
of  the  fall  in  feet.  If  a  certain  body,  say  a  cannon  ball,  falls  from  a  height 
of  750  feet,  what  will  be  its  velocity  when  it  strikes  the  ground? 

Solution.— The  square  root  of  750  is  27.3864-,  and  8.02  X  27.386 
=  219.64—  feet  per  second,  or  about  2.5  miles  per  minute.     Ans. 

Example  3. — The  intensity  of  light  varies  inversely  as  the  square  of 
the  distance  of  the  light  from  the  object  illuminated.  If  the  intensity  of 
illumination  of  a  certain  object  5  feet  from  the  source  of  light  is  32  candle- 
power,  at  what  distance  will  the  intensity  of  illumination  be  IS  candlepower? 

Solution. — Expressed  as  a  direct  proportion,  32  :  18  =  52  :  x2.  It  is 
necessary  to  square  the  distances  5  and  x,  because  the  intensity  varies  in- 
versely as  the  square  of  the  distance.     Inverting  the  terms  in  the  second 

32  X  25 
ratio,  32  :  18  =  x2  :  52,  from  which  x2  =  jg —    =    44.4444+.     Now     if 

x2  =  44.4444,  x  must  equal  the  square'root  of  44.4444,  that  isx  =  \/44.4444 
=  6.6667—  feet.  Ans.  This  result  might  also  have  been  obtained  as 
follows:   32X26  _  400  ^  ^400  .  |  .   8,   .   e.0607-    feet.     The 

above  result  means  this:  if  an  object  is  illuminated  with  an  intensity  of  32 
candlepower  when  5  feet  from  the  light,  it  will  lie  illuminated  with  an  in- 
tensity of  only  18  candlepower  when  6f  feet  from  the  light. 

Example  4. — The  strength  of  a  simple  beam  (one  that  is  merely  sup- 
ported at  its  ends)  varies  directly  as  its  breadth,  directly  as  the  square  of  its 
depth,  and  inversely  as  its  length.  If  a  simple  beam  made  of  hemlock  is 
20  feet  long,  has  a  breadth  of  3  inches,  a  depth  of  8  inches,  and  will  safely 
carry  a  load  of  960  pounds,  uniformly  distributed  over  the  beam,  what  must 
be  the  depth  of  a  similar  beam  having  a  length  of  Ki  feet  and  a  breadth  of  2 
inches  to  carry  safely  a  uniform  load  of  1250  pounds? 

Solution. — This  is  evidently  a  problem  in  compound  proportion.     Stated 
as  a  direct  proportion,  3  X  82  X  20  :  2  X  x2  X  16  =  960  :  1250.     It  is  neces- 
sary to  square  8  and  x,  because  the  strength  varies  as  the  square ol  the  depth 
7 


98  ARITHMETIC  §1 

But  the  strength  varies  inversely  as  the  length;  hence,  transposing  20  and 
16,    the    lengths,    3  X  S2  X  16  :  2  X  x2  X  20  =  960  :  1250,    from    which 

X2  =  sj^x^ie^ypo  =  100    But  if  x2  =  100  x  =  v1q6  =  10;  henCG) 

the  depth  of  the  beam  must  be  10  inches.     Ans. 

The  problems  just  given  will  suffice  to  show  the  importance  of 
knowing  some  method  of  extracting  the  square  root  of  numbers. 


PERCENTAGE 

165.  It  is  frequently  desirable  to  compare  numbers  or 
quantities  not  in  relation  to  each  other,  but  in  relation  to  some 
fixed  number,  called  the  base.  For  example,  the  sum  of  7,  6, 
and  12  is  25.  Now  to  compare  7,  6,  and  12  with  25,  form  the 
ratios  7  :  25,  6  :  25,  and  12  :  25.  The  values  of  these  ratios  are 
found  by  dividing  the  first  terms  by  the  second,  thus  obtaining 
775  =  .28,  -/ff  =  -24,  and  |f  =  .48.  Note  that  the  sum  of  the 
fractions  is  To  +  To  +  if  =  1,  and  the  sum  of  the  decimal 
equivalents  of  the  fractions  is  .28  +  .24  +  .48  =  1.00  =  1,  also, 
as  it  must.  In  the  case  of  the  ratios,  the  base  is  25;  but  in  the 
case  of  the  fractions  and  decimals,  the  base  is  1,  a  much  more 
convenient  number,  and  one  that  does  not  change,  as  will 
continually  be  the  case  with  ratios. 

In  business  transactions,  and  in  many  other  cases  that  arise  in 
practice,  it  is  convenient  to  have  TiT  =  .01  as  the  base,  thus  making 
the  values  of  the  ratios  100  times  as  large  as  those  obtained  above. 
With  .01  as  the  base,  the  values  of  the  above  ratios  become 
•28  -h  rU  =  -28  X  100  =  28,  .24  X  100  =  24,  .48  X  100  =  48, 
and  28  +  24  +  48  =  100.  In  the  last  case,  the  three  numbers 
added  are  called  28  per  cent,  24  per  cent,  and  48  per  cent, 
respectively,  of  25,  which  is  100  per  cent  of  25. 

The  term  per  cent  is  an  abbreviation  of  the  Latin  words  per 
centum,  which  mean  by  the  hundred,  and  percentage  means  a  cer- 
tain number  of  hundredths  of  some  number.  Thus,  28  per  cent 
of  25  is  25  X  r0V  =  25  X  .28  =  7,  and  7  is  the  percentage 
obtained  by  multiplying  25  by  28  hundredths.  Also,  7  per  cent 
of  438  is  438  X  jfo  =  438  X  .07  =  30.66,  and  30.66  is  the 
percentage  obtained  by  multiplying  438  by  7  hundredths. 

166.  The  sign  or  symbol  that  is  used  to  denote  per  cent  is%; 
thus,  7r;  is  read  7  per  cent,  11J4%  is  read  11*4  per  cent,  etc. 
The  number  that  is  placed  before  the  symbol  and  shows  how 


PERCENTAGE 


99 


many  hundredths  are  to  be  taken  or  considered  is  called  the 
rate  per  cent;  thus,  in  the  last  sentence,  7  and  11^4  :il,>  rates  per 
cent.  When  the  rate  per  cent  is  expressed  decimally  or  frac- 
tionally, thus  denoting  its  actual  value,  it  is  called  the  rate.     For 

5 
instance,  if  the  rate  per  cent  is  5%,  the  rate  is  .05  or  ..    •  ii  the 


rate  per  cent  is  58.7%,  the  rate  is  .587  or 


58.7         :»s7 


,  etc. 


100        1000 

The  rate,  therefore,  is  always  equal  to  the  rate  per  cent  divided  by 
100;  and  the  rate  per  cent  is  100  times  the  rate. 

167.  The  rate  can  frequently  be  expressed  as  a  simple  fraction 
by  reducing  the  common-fraction  form  of  the  rate  to  its  lowest 

J>L        26         ,  .,„,       6.25 

100  "  400  ~  ir"  01   °4/o  "   100 

625 

=       "       =  ^V-     In  the  table  below,  some  of  the  rates  per  cent 

commonly  used,  with  their  decimal  and  common  fraction  equiva- 
lents, are  given. 


terms.     For  instance,  6|% 


Rate  Per  Cent 

Rate 

Kate  Per  Cent 

Rate 

li 

.015   =.;10 

25 

.25  =  \ 

2 

•  02   =6v 

33j 

■  OO  3  —   3 

3* 

031   =  ,<„ 

37  £ 

.375=  f 

4 

.04   =^g 

40 

.4    =   g 

4£ 

.041   „A 

50 

5   =  J 

5 

.05   =  2\, 

62  \ 

.625=  f 

61 

.0625  =A 

66  § 

66|=  | 

8 

.08   =  .,*, 

75 

.75  =  J 

10 

•1 

83j 

•833=  g 

m 

.125   =| 

87  i 

.875 

16| 

I65   =1 

100 

1    =1 

20 

•2    =1 

250 

2.5   =21 

i 

8 

.00125  =  ,!,„ 

475 

4.75 

168.  The  majority,  if  not  all,  of  the  fractions  in  the  above  table 
may  be  used  to  advantage  in  computing  the  percentage  instead 
of  their  decimal  equivalents.  For  example,  to  find  12^  %  of 
35.164,  simply  divide  35.164  by  8,  obtaining  4.3955;  this  is  cor- 
rect, since  12^%  =  £,  and  35.164  X|  =  35.164  +  8.  Simi- 
larly, for  the  same  reasons,  83>$%  of  439.2  =  439.2  =  % 
_  ■±o^!_^_o  _  366      It  .g  evidentiy  casicr  to  multiply  by  the 


100  ARITHMETIC  §1 

fractions  than  it  is  to  multiply  by  their  decimal  equivalents,  and 

there  is  less  liability  of  making  mistakes. 

169.  The  base  may  now  be  denned  as  the  number  that  is 
multiplied  by  the  rate  to  obtain  the  percentage.  In  the  first 
case  of  Art.  168,  35.164  is  the  base  and  4.3955  is  the  percentage; 
in  the  second  case,  439.2  is  the  base  and  366  is  the  percentage. 
Consequently. 

Rule. — To  find  the  percentage,  multiply  the  base  by  the  rate. 

Example. — Out  of  a  lot  of  1500  roams  of  paper,  37%  were  sold  the  first 
month  and  439c  were  sold  the  second  month;  how  many  were  sold  in  the  two 
months? 

Solution. — This  problem  may  be  solved  in  two  ways:  1st  method.  The 
number  of  reams  sold  the  first  montn  is  1500  X  .37  =  555;  number  sold  the 
second  month  is  1500  X  .43  =  G45;  number  sold  in  the  two  months  is  then 
555+645  =  1200.  .Ins.  2d  method.  If  37  %  were  sold  the  first  month  and 
43%  were  sold  the  second  month,  37%  +  43%  =  80%  were  sold  in  the  two 
months;  and  1500  X  .80  =  1200  reams  were  sold  in  the  two  months.     Ans. 

170.  If  the  product  of  the  base  and  the  rate  equals  the  percent- 
age, the  rate  must  equal  the  percentage  divided  by  the  base,  since 
the  product  of  two  factors  divided  by  one  of  the  factors  must  give 
the  other  factor  for  the  quotient.     Therefore, 

Rule. — To  find  the  rate,  divide  the  percentage  by  the  base.  The 
rate  per  cent  is  100  times  the  rate. 

Example. — Out  of  a  lot  of  1500  reams  of  paper,  555  were  sold;  what  per 
cent  of  the  paper  was  sold? 

Solution. — Here  555  is  the  percentage  and  1500  is  the  base,  since  it  is  the 
number  which,  multiplied  by  some  rate,  gives  a  product  of  555.  Hence, 
555  -5-  1500  =  .37  =  the  rate,  and  .37  X  100  =  37,  the  rate  per  cent. 
Therefore,  the  per  cent  of  paper  sold  was  37%.     Ans. 

171.  If  the  rate  and  percentage  are  known  and  it  is  desired  to 
find  the  base,  divide  the  percentage  by  the  rate.  This  is  evi- 
dently correct,  since,  if  the  product  of  the  base  and  rate  equals 
the  percentage,  the  percentage  divided  by  the  rate  must  equal  the 
base.     Consequently, 

Rule. —  To  find  the  base,  divide  the  percentage  by  the  rate. 

Example. — In  a  certain  month,  43%  of  a  certain  lot  of  pulp  was  sold; 
if  the  number  of  bales  sold  was  645,  how  many  bales  were  in  the  lot? 

S<  .lution. — Here  the  rate  is  .43  and  the  percentage  is  645;  hence,  the  base 
is  645  -S-  .43  =  1500,  the  number  of  bales  in  the  lot.     Ans. 

172.  It  is  to  be  noted  that  the  rate  per  cent  and  the  percentage 
really  represent  the  same  thing.  The  rate  per  cent  is  the  number 
of  parts  in  100  parts  of  the  base,  while  the  percentage  is  the  actual 


§1  PERCENTAGE  101 

number  of  parts  of  the  base.  In  other  words,  the  rate  per  cent 
is  found  on  the  assumption  that  the  base  is  always  100,  while 
the  percentage  is  computed  on  the  actual  value  of  the  base. 

173.  There  are  two  other  terms  used  in  percentage — the  amount 
and  the  difference.  The  amount  is  equal  to  the  sum  of  the  base 
and  percentage,  the  difference  is  equal  to  the  remainder  obtained 
by  subtracting  the  percentage  from  the  base.  To  illustrate  the 
meaning  of  these  terms,  suppose  that  the  price  of  alum  is  $42.00 
per  ton,  and  that  the  price  was  increased  12^%;  what  is  the  new 
price?  $42  +  8  =  $5.25  =  12>£%  of  $42.00;  then  the  new  price 
is  evidently  $42.00  +  $5.25  =  $47.25  =  the  amount,  since  it  is 
equal  to  the  sum  of  the  base  ($42.00)  and  the  percentage  ($5.25). 
Again,  suppose,  as  before,  that  the  price  is  $42.00,  but  that  the 
price  is  reduced  \2Yi%,  what  is  the  new  price?  Since  12^%  of 
$42.00  is  $5.25,  the  new  price  is  $42.00  -  $5.25  =  $36.75  =  the 
difference,  since  it  is  equal  to  the  base  minus  the  percentage. 

174.  The  amount  may  be  found  in  an  easier  way.  Since  the 
base  always  represents  1  or  100%,  the  amount  may  be  represented 
by  1  +  rate  or  by  100%  +  rate  per  cent;  hence,  if  the  base  and 
rate  are  given,  the  amount  may  be  found  by  multiplying  the 
base  by  1  +  rate.  Thus,  referring  to  Art.  173,  the  new  price 
after  the  increase  is  $42  X  (1  +  .125)  =  $42  X  1.125  =  $47.25, 
since  12>£%  =  .125.     Therefore, 

Rule. — The  amount  equals  the  base  multiplied  by  1  plus  the  rate. 

Example. — If  a  man's  wages  is  $3.20  per  day,  and  ho  receives  an  increase 
of  15%,  how  much  does  he  then  receive  per  day? 

Solution. — The  base  is  $3.20,  the  rate  is  .15,  and  the  new  rate  of  wages 
is  the  amount,  which  is  $3.20  X  1.15  =  $3.68.     Ans. 

175.  If  the  amount  and  the  rate  are  known,  and  it  is  desired  to 
find  the  base  on  which  the  rate  was  computed,  divide  the  amount 
by  1  plus  the  rate.  This  is  evidently  correct,  since  the  amount  is 
equal  to  the  base  multiplied  by  1  plus  the  rate.     Hence, 

Rule. — To  find  the  base,  divide  the  amount  by  1  plus  the  rate. 

Example. — If  a  man  receives  $3.GS  cents  per  day  in  wages,  and  this  is 
15%  more  than  he  formerly  received,  how  much  did  he  formerly  receive'' 

Solution. — Here  $3.08  is  some  number  plus  l.v,  of  that  number;  that 
is,  $3.68  is  the  amount  and  15%  is  the  rate  per  cent.  Therefore,  he  formerly 
received  $3.68  -r-  1.15  =  $3.20.     Ans. 

176.  If  the  base  and  amount  are  given  and  it  is  desired  to  know 
the  rate,  divide  the  amount  by  the  base  and  subtract  1  from  the 
quotient.     This  is  evidently  correct,  since  dividing  the  amount 


102  ARITHMETIC  §1 

by  the  base  gives  a  quotient  that  equals  1  plus  the  rate,  and  sub- 
tracting 1  from  this  leaves  the  rate. 

Rule. — To  find  the  rale  when  the  base  and  amount  are  known, 
divide  the  amount  by  the  base  and  subtract  1  from  the  quotient. 

Example. — If  a  man's  wages  are  S3. 68  per  day  and  he  formerly  received 
S3. 20  per  day,  what  rate  per  cent,  increase  did  he  receive? 

Solution. — Here  S3. 68  is  the  amount,  since  it  equals  S3.20  plus  a  certain 
percentage  of  S3. 20,  the  base.  Consequently,  3.68  4-  3.20  =  1.15;  1.15  — 
1  =  .15,  the  rate;  and  .15  X  100  =  15%-  Ans.  The  solution  may  also  be 
obtained  as  follows:  S3. 68  —  S3. 20  =  S0.48,  the  actual  increase  per  day  that 
he  received,  and  which  is  the  percentage.  Hence,  the  rate  is  .48  -5-  3.20 
=  .15,  and  the  rate  per  cent  is  .15  X  100  =  15%.     Therefore,#also, 

Rule. — To  find  the  rate  when  the  base  and  the  amount  are  known, 
subtract  the  base  from  the  amount  and  divide  the  remainder  by  the  base. 

The  second  rule  is  rather  easier  to  remember  and  apply  than 
the  first  rule. 

177.  The  rules  just  given  concerning  the  amount  may  also  be 
used  when  the  difference  is  given,  by  subtracting  the  rate  instead 
of  adding  it.  Thus,  referring  to  Art.  173,  the  new  price  after  the 
old  price  had  been  reduced  12>£%  is  $42  X  (1  -  .125)  =  $42  X 
.875  =  $36.75.     Therefore, 

Rule. —  To  find  the  difference,  multiply  the  base  by  1  minus  the 
rate. 

In  connection  with  problems  relating  to  money  and  prices,  the 
amount  deducted  from  a  fixed  price  or  amount  (the  base)  is 
called  the  discount.  Insofar  as  percentage  is  concerned,  the 
words  percentage  and  discount  mean  the  same  thing,  when  finding 
the  difference. 

Example. — If  a  certain  article  is  priced  at  $4.75  and  it  is  sold  at  a  discount 
of  8%,  how  much  was  received  for  it? 

Solution. — The  base  is  S4.75,  the  rate  is  .08,  and  the  difference  is  re- 
quired. Applying  the  rule,  S4.75  X  (1  -  .08)  =  S4.75  X  .92  =  S4.37. 
Ans.  The  solution  may  also  be  obtained  as  follows:  $4.75  X  .08  =  $0.38, 
the  discount;  $4.75  -  $0.38  =  $4.37.  Ans.  Either  method  may  be  used, 
whichever  appears  to  be  easier. 

178.  If  the  difference  and  rate  are  known,  and  it  is  desired  to 
find  the  base,  divide  the  difference  by  1  minus  the  rate.  For, 
since  the  product  of  the  base  and  1  minus  the  rate  equals  the 
difference,  the  base  must  equal  the  difference  divided  by  1  minus 
the  rate.     Consequently, 

Rule. — To  find  the  base  when  the  rate  and  difference  are  known, 
divide  the  difference  by  1  minus  the  rate. 


§1  PERCENTAGE  103 

Example. — $4.37  was  paid  for  a  certain  article  that  was  bought  at  a  dis- 
count of  8%;  what  was  the  original  price  of  the  article? 

Solution. — Here  $4.37  is  the  difference  and  .08  is  the  rate;  the  base  = 
the  original  price  =  $4.37  -5-  (1  -  .08)  =  $4.37  -r-  .92  =  $4.75.     Ans. 

179.  If  the  difference  and  base  are  known,  and  it  is  desired  to  find 
the  rate,  subtract  the  difference  from  the  base  and  the  remainder 
will  be  the  discount;  divide  the  discount  by  the  base,  and  the 
quotient  will  be  the  rate. 

Example. — If  an  article  is  bought  for  $4.37  and  the  original  price  was 
$4.75,  what  was  the  per  cent  of  discount  received? 

Solution. — The  actual  discount  received  is  $4.75  -  $4.37  =  $0.38,  and 
$4.75  is  the  base;  hence,  0.38  *  4.75  =  .08  =8%  of  the  original  price  = 
per  cent  of  discount.     Ans. 

180.  Chain  Discount. — Certain  manufacturers  that  deal  in 
goods  that  are  subject  to  rapid  changes  in  prices  issue  cataloges 
in  which  the  prices  are  quoted  at  a  much  higher  figure  than  they 
think  will  ever  be  asked.  They  then  issue  what  are  called  dis- 
count sheets,  which  are  quickly  printed,  and  are  subject  to  change, 
in  accordance  with  the  state  of  the  market.  When  more  than 
one  discount  is  quoted  on  any  one  article,  it  is  called  a  chain 
discount.  In  computing  a  chain  discount,  that  quoted  first  is 
deducted  from  the  list  price,  the  second  discount  is  deducted  from 
the  remainder,  the  third  discount  from  the  last  remainder,  etc. 
It  is  never  allowable  to  add  the  rates  of  discount  and  then  deduct. 

As  an  example,  suppose  that  the  price  (catalog)  of  a  certain 
article  is  $26.60,  and  that  discounts  of  60,10  and  5%  are  offered; 
what  is  the  real  price  of  the  article? 

$26.60  X  (1  -  .60)  =  $10.64,  the  price  after  60%  has  been 
deducted.  $10.64  X  (1  -  .10)  =  $9,576,  or  $9.58,  the  price 
after  deducting  60  and  10%.  $9.58  X  (1  -  .05)  =  $9,101  or 
$9.10,  the  price  after  all  the  discounts  have  been  taken  out. 

This  same  result  may  be  obtained  by  multiplying  $26.60  by 
the  continued  product  of  1  minus  the  different  discounts,  that  is 
by  (1  -  .60)  X  (1  -  .10)  X  (1  -  .05)  =  .40  X  .90  X  .95  = 
.342;  thus,  $26.60  X  .342  =  $9.0972,  or  $9.10,  as  before.  The 
real  discount  in  this  case  is,  therefore,  100  —  34.2  =  65.8%, 
which  may  be  called  the  equivalent  discount,  and  1  —  .342  = 
.658  may  be  called  the  equivalent  discount  rate. 

Rule. —  To  find  the  equivalent  discount  rate  for  any  chain  dis- 
count, subtract  all  the  discount  rates  from  1  and  find  their  product; 
then  subtract  this  product  from  1.  The  equivalent  discount  equals 
the  equivalent  discount  rate  multiplied  by  100.     The  net  price  or 


104  ARITHMETIC  §1 

nd  cost  may  be  obtained  by  subtracting  all  the  discount  rates  from  1, 
finding  their  product,  which  multiply  by  the  given  -price  or  cost. 

Example. — A  bill  of  goods  amounting  to  $102.0-1  is  subject  to  a  chain 
discount  of  S7>£,  10,  5,  and  2%%;  what  is  the  equivalent  discount  per  cent 
and  how  much  must  be  paid  to  settle  the  bill? 

Solution. — The  equivalent  discount  rate  is  1  —  (1.  —  .875)  X  (1  —  10) 
X  (1  -  .05)  X  (1  -  .025)  =  1  -  .125  X  .9  X  .95  X  .975  =  1  -  .104203125 
=  .895796875,  or  89.5796875%.  Ans.  The  amount  to  be  paid  to  settle 
the  bill  is  $102.04  X  .104203125  =  $10.63+.     Ans. 

181.  Gain  or  Loss  Per  Cent. — If  an  article  is  bought  for  a 
certain  amount  and  is  sold  for  another  amount,  there  is  a  gain  or 
loss  equal  to  the  difference  between  the  two  amounts.  To  find 
the  g;iin  or  loss  per  cent,  divide  the  gain  or  loss  by  the  cost.  Thus, 
if  a  manufacturer  finds  that  it  costs  him  $2.43  to  turn  out  a 
certain  article,  and  he  sells  it  for  $3.15,  he  gains  $0.72,  and  the 
gain  per  cent  is  .72  +  2.43  =  .2963;  .2963  X  100  =  29.63%. 
But  if  he  had  sold  it  for  $2.25,  he  would  have  lost  $2.43  -  $2.25 
=  $0.18,  and  the  loss  per  cent.  =  .18  -^  2.43  =  .0741;.  0741 
X  100  =  7.41%. 

Again,  suppose  that  the  transmission  system  of  a  certain  power 
plant  is  found  to  absorb  4.35  horsepower,  but  after  making  certain 
changes,  it  absorbs  only  3.08  horsepower;  what  is  the  saving  in 
per  cent  of  horsepower  wasted  in  transmission?  The  saving  in 
horsepower  is  4.35  —  3.08  =  1.27;  the  horsepower  wasted  before 
the  change  is  the  base  in  computing  the  per  cent;  hence,  1.27  -S- 
4.35  =  .292-  ;  and  .292  X  100  =  29.2%.  In  other  words,  29.2% 
of  the  power  formerly  wasted  is  saved  by  the  change. 

In  calculations  of  this  kind,  it  is  sometimes  a  little  difficult  to 
determine  which  of  the  two  given  numbers  is  the  base;  this  can 
always  be  determined  correctly  by  remembering  that  a  gain  or 
loss  involves  a  change  of  some  kind;  in  the  case  of  buying  and 
selling,  the  selling  is  a  change  (of  ownership) ;  in  the  case  of  the 
power  plant,  4.35  horsepower  was  changed  to  3.08  horsepower. 
The  number  used  to  divide  the  gain  or  loss  is  the  number  that  is 
changed;  in  the  first  illustration  above,  the  number  changed  was 
$2.43,  and  in  the  second  illustration  it  was  4.35. 

Rule. — To  find  the  gain  or  loss  per  cent,  divide  the  gain  or  loss 
by  the  number  that  is  changed  and  multiply  the  quotient  by  100. 

Example  1. — Suppose  the  indicated  horsepower  of  a  certain  non-condens- 
ing steam  engine  is  168,  and  that  it  is  20 4  after  a  condenser  has  been  attached. 
What  is  the  gain  per  cent  in  horsepower'.' 

Solution. — The  horsepower  before  the  change  was  168,  and  after  the 


§1  PERCENTAGE  105 

change  it  was  204;  the  gain  is  204  -  168  =  3G;  30  -^  168  =  .2143,  and 
.2143  X  100  =  21.43%,  the  gain  per  cent.     Ans. 

Example  2. — A  certain  business  is  valued  at  $234,517.00;  three  years 
later,  it  is  valued  at  $187,250;  what  was  the  depreciation  in  per  cent? 

Solution.— The  actual  depreciation  was  $234,517  -  $187,250  =  $47,267; 
the  value  before  the  change  was  $234,517;  47,267  +  234,517  =  .2016-; 
hence,  the  depreciation  in  per  cent  was  .2016  X  100  =  20.16%.     Ans. 


EXAMPLES 

(1)  What  is  4%%  of  $683.32  to  the  nearest  cent?  Ajis.  $32.46. 

(2)  If  475  is  62>£%  of  some  number,  what  is  the  number?         Ans.  760. 

(3)  What  is  %%  of  $4521?  Ans.  $16.95. 

(4)  What  per  cent  of  840  is  119?  Ans.  14K  %• 

(5)  If  an  automobile  was  bought  for  $1350  and  after  being  used  for  a  year 
was  sold  for  $925,  what  was  the  loss  per  cent?  Ans.  31.48%. 

(6)  Bought  some  dye  stuff  for  $21.85,  receiving  a  discount  of  5%  from  the 
selling  price;  what  was  the  selling  price?  Ans.  $23.00. 

(7)  A  machinist  turned  47  bolts  in  one  day,  which  was  almost  7%  more 
than  he  turned  the  day  before;  how  many  did  he  turn  the  day  before? 

Ans.  44. 

(9)  Bought  a  line  of  supplies,  the  bill  amounting  to  $428.73  and  subject 
to  a  discount  of  33%,  10,  and  3%  if  settled  within  ten  days;  how  much  is 
required  to  settle  in  ten  days?  Ans.  $249.52. 

(10)  A  test  showed  that  the  coal  consumption  in  the  boiler  used  to  furnish 
steam  for  a  non-condensing  engine  was  660  pounds  per  hour;  after  attach- 
ing a  condenser,  only  541  pounds  of  coal  per  hour  were  required.  What  was 
the  saving  per  cent?  Ans.  18  +  %. 

(11)  The  average  daily  production  of  a  paper  mill  in  1917  was  94  tons; 
in  1918,  it  was  105  tons;  what  was  the  per  cent  of  increase?    Ans.  11.7%. 

(12)  A  mill  was  using  1.4  cords  of  wood  per  ton  of  paper;  as  the  result  of 
certain  changes,  a  saving  of  4%  of  wood  was  effected.  What  amount  of 
wood  was  then  used  per  ton?  Ans.  1.344  cords  per  ton. 

(13)  A  pulp  mill  increased  its  production  43%,  producing  160  tons  per 
day;  how  much  did  it  formerly  produce?  Ans.  112  tons  per  day. 

(14)  If  165  pounds  of  clay  is  used  in  an  order  of  2240  pounds  of  paper, 
what  per  cent  of  the  finished  product  is  clay?  Ans.  7.37%. 

(15)  A  sample  of  paper  weighs  9.372  grams,  of  which  0.621  grams  arc 
water  and  8.751  grams  are  fiber;  what  is  the  percentage  of  water  and  fiber 
in  the  paper  expressed  as  per  cent?  Ans.  /  6.63%    of    water. 

\  93.37%    of    fiber. 

(16)  A  sample  of  coal  shows  11.3%  ash;  assuming  that  all  the  combustible 
part  of  the  coal  is  consumed,  how  many  pounds  of  ashes  must  be  handled 
per  ton  of  2240  pounds  of  coal?  Ans.  253.12  pounds. 

(17)  After  having  been  in  use  for  some  time,  it  was  found  that  a  58-foot 
belt  had  stretched  3%;  what  was  the  length  of  the  belt  after  being  stretched? 

Ans.  59.74  feet. 


106  ARITHMETIC  §1 

(18)  A  paper  shrinks  l.tiL",  from  wet  end  to  calenders;  if  it  was  164 
inches  wide  cm  tin-  wire,  how  wide  is  it  when  dry'.  Ans.  161.34  inches. 

(19)  How  much  bone  dry  fiber  is  contained  in  793  tons  of  wet  pulp,  the 
moisture  content  being  A'.V ,  '.'  Ans.  452  tons. 

(20)  The  bone  dry  weight  of  pulp  is  90%  of  the  air  dry  weight,  which  is 
the  basis  of  payment.  Referring  to  example  19,  what  is  the  equivalent  air 
dry  pulp?  Ans.  502.22  tons. 

COMPOUND  NUMBERS 

182.  A  compound  number  is  one  that  requires  more  than  one 
unit  to  express  it ;  thus,  if  the  length  of  a  piece  of  hose  is  stated  to 
be  12  feet  7  inches,  12  feet  7  inches  is  a  compound  number,  because 
two  units — the  foot  and  the  inch — are  required  to  express  the 
length.  Had  the  length  been  stated  as  12  ]^2  feet,  the  number 
would  be  called  a  simple  number,  since  only  one  unit — the  foot — 
is  required  to  express  it. 

Compound  numbers  are  usually  denominate  numbers,  a 
denominate  number  being  one  having  its  unit  or  units  de- 
nominated, or  named. 

183.  A  denominate  number  is  always  a  concrete  number  (see 
Art.  7),  and  when  only  one  unit  is  named,  as  5  yards,  8  pounds, 
etc.,  it  is  a  simple  number.  An  abstract  number  is  usually  a 
simple  number,  but  it  may  be  treated  as  a  compound  denominate 
number  if  desired.  Thus,  364  may  be  regarded  as  3  hundreds  6 
tens  4  units. 

184.  In  the  case  of  abstract  numbers,  each  figure  belongs  to  a 
denomination  that  is  ten  times  as  large  as  the  denomination  of 
the  next  figure  on  its  right  and  one-tenth  as  large  as  the  denomi- 
nation of  the  next  figure  on  its  left.  In  connection  with  de- 
nominate numbers,  however,  this  is  seldom  or  never  the  case. 
For  instance,  in  the  compound  denominate  number  4  yards  2 
feel  8  inches,  it  takes  12  inches  to  make  1  foot,  the  next  higher 
denomination,  and  it  takes  3  feet  to  make  1  yard,  the  next 
denomination  higher  than  feet.  For  this  reason,  it  is  necessary 
to  memorize  certain  tables  showing  the  relation  of  the  different 
denominations  for  different  compound  numbers. 

In  order  to  save  time  and  space  in  writing,  the  names  of  the 
units  arc  abbreviated,  and  these  abbreviations  are  given  in  the 
tables  that  follow. 

185.  The  t  ablcs  that  follow  should  all  be  thoroughly  committed 
to  memory.     Only  the  tables  in  common  use  are  given  here. 


§1  COMPOUND  NUMBERS  107 

Under  each  table  is  a  subsidiary  table  thai  shows  the  relation 
between  the  different  units;  it  is  not  necessary  to  memorize  these, 
although  it  will  be  found  convenient  to  remember  some  of  the 
principal  equivalents,  as,  for  example,  that  5280  feet  make  a 
mile,  that  36  inches  make  a  yard,  etc. 

TABLE  I 

LINKAR  MEASURE 

12  inches  (in.) =1  foot ft. 

3  feet =    l  yard yd. 

h%.  yards =    l  rod rd. 

40  rods =1  furlong fur. 

8  furlongs  (320  rd.) =    1  mile mi. 


in. 


mi.         fur.  rd.  yd.  ft, 

1     -     8     -     320     -     1760     =     5280  =  63,360 

1      =       40      =       220      =       660  =  7,920 

1      =       5.5      =      16.5  =  198 

1      =           3  =  36 

1  =  12 

186.  Another  abbreviation  for  feet  and  inches,  much  used  by 
draftsmen,  mechanics,  etc.  is  (')  for  feet  and  (")  for  inches;  hence, 
4  feet  8  inches  may  be  written  either  4  ft.  8  in.  or  4'  8".  When 
the  latter  form  is  used,  it  is  advisable  to  place  a  dash  between  the 
feet  and  inches;  thus,  4' — 8".  The  subsidiary  table  is  useful  in 
expressing  higher  units  in  terms  of  the  lower  units,  and  vice  versa. 
For  example,  if  it  were  desired  to  express  37  rd.  in  inches,  the 
table  shows  that  1  rd.  =  198  inches;  hence,  37  rd.  =  37  X  198 
=  7326  in.     Again,  what  fraction  of  a  mile  is  1360  feet?     From 

the  table,  1  mi.  =  5280  ft.;  hence,  1360  ft.  =  1^  =    \\   mi. 

=  .2576  -  mi. 

TABLE  II 

SQUARE  MEASURE 

144  square  inches  (sq.  in.).. .     =    1  square  foot sq.    ft. 

9  square  feet =    1  square  yard sq.  yd. 

30 >4  square  yards =    1  square  rod sq.    rd. 

160  square  rods =    1  acre V. 

640  acres =    1  square  mile sq.  mi. 

sq.  mi.  A.         sq.  rd.  Bq.  yd.  sq.  ft.  sq.  in. 

1  =  640  =  102,400  =  3,097,600  =  27,878,  MM)  =  4,014,489,600 

1  =  160  =  4,840  =         43,560  =  6,272,640 

1  =  30.25=         272.25  =  39,204 

1  =  9  =  1,296 

1  =  144 


108  ARITHMETIC  §1 

187.  A  plot  of  ground  in  the  form  of  a  square,  each  side  of  which 
measures  208.71  feet,  say  208  ft.  9  in.,  contains  one  acre. 

TABLE  III 

CUBIC  MEASURE 

1728  cubic  inches  (cu.  in.)...  .    =    1  cubic  foot cu.  ft. 

27  cubic  feet =   1  cubic  yard cu.  yd. 

128  cubic  feet =  1  cord  (wood) cd. 

24%  cubic  feet =  1  perch  (stone,  masonry)  P. 

cu.  yd.  cu.  ft.  cu.  in. 

1  =         27  46,656 

1  =  1,728 

188.  The  cord  is  used  onl}-  in  measuring  wood.  A  pile  of  wyood 
8  ft.  long,  4  ft.  wide,  and  4  ft.  high  contains  one  cord.  The  perch 
is  used  in  measuring  stone  and  brick  walls  and  other  masonry. 
Some  contractors  allow  25  cubic  feet  to  the  perch,  but  24  % 
cubic  feet  is  the  correct  value. 

TABLE  IV 

AVOIRDUPOIS  WEIGHT 

437^  grains  (gr.) =    1  ounce oz. 

16      ounces =    1  pound lb. 

100      pounds 

20    hundredweight  (2000  lb.) 

T.  cwt.  lb. 

1      =     20      =     2000      = 

1      =        100      = 

1      = 


189.  The  ton  of  2000  pound,  called  the  short  ton,  is  the  one 
commonly  used.  The  ton  of  2240  pounds  is  called  the  long  ton, 
and  is  used  to  weigh  coal,  pig  iron,  and  other  coarse  commodities; 
it  is  the  basis  of  freight  rates  on  foreign  exports.  In  connection 
with  the  long  ton,  14  pounds  make  a  stone,  2  stones  make  a 
quarter,  4  quarters  make  a  hundredweight,  and  20  hundred- 
weight make  a  ton;  hence. 


=    1  hundrec 

[weight....  cwt. 

=    1  ton .... 

T. 

oz. 

gr. 

32,000      = 

14,000,000 

1,(100      = 

700,000 

L6      = 

7,000 

1      = 

437.5 

LONG 

TON 

T. 

cwt. 

qr. 

St. 

lb. 

1 

20 

= 

80 

= 

160 

= 

2240 

1 

= 

4 

= 

8 

= 

112 

1 

= 

2 
1 

= 

28 
14 

compoi'N])  ni:mbi;rs 


109 


190.  For  measuring  medicines,  jewelry,  gold,  silver,  etc.,  the 
Troy  ounce,  which  contains  480  grains  is  used.  The  Troy  pound 
contains  12  Troy  ounces  or  5760  grains.  The  Troy  pound  is 
therefore,  only  i i-Jftft  =  4?lth  of  an  avoirdupois  pound.  A  pound 
of  gold  thus  weighs  5760  -f-  437.5  =  13iVs  =  13.1657  +  avoir- 
dupois ounces. 


TABLE  V. 

LIQUID  MEASIKF, 


4      gills  (gi.) 

2      pints 

4      quarts 

313^  gallons 

2      barrels  (03  gallons). 


hhd. 

1 


bbl. 
2 

1 


1  pint pt. 

1  quart qt. 

1  gallon gal. 

1  barrel bbl. 

1  hogshead hhd 


gal. 
63 

31. 

1 


qt. 

252 

126 

4 

1 


pt. 
504 
252 


2010 

1008 

32 

8 
4 


191.  The  United  States,  or  wine,  gallon  contains  231  cubic 
inches,  and  a  gallon  of  water  weighs  very  nearly  8f  pounds.  A 
cubic  foot  contains  1728  ■*■  231  =  7.481  gallons,  or,  roughly,  1\ 
gallons.  The  British  imperial  gallon,  used  in  Great  Britain  and 
Canada,  contains  277.463  cubic  inches,  and  a  gallon  of  water  at 
a  temperature  of  62  degrees  Fahrenheit  weighs  exactly  10  pounds. 
The  British  imperial  gallon  is  equal  to  1.2  U.  S.  gallons,  very 
nearly. 

What  is  known  as  the  fluid  ounce  is  the  weight  of  TVth  of  a 
pint  of  water,  or  listh  of  a  gallon  of  water.  Since  a  gallon  of 
water  weighs  8}i  pounds,  a  fluid  ounce    weighs  8 J  X  16  -f-  128 

2.5. 


=  I?1*  =  if  ounces 


2  pints  (.pt.). 
8  quarts. . .  . 
4  pecks 


TABLE  VI 

UHY  MEASURE 


bu. 
1 


Pk. 
=     4 

1 


1  quart qt. 

1  peck pk. 

1  bushel bu. 

qt.  pt. 

32     =      M 

8     =       1»- 

1     =        2 


192.  The  unit  of  dry  measure  is  the  Winchester  bushel,  which 
contains  2150.42  cubic  inches;  hence,  the  dry   quart    contains 


110 


ARITHMETIC 


§1 


cir. 

quad.  (L) 

deg.  (°) 

1 

4 

360 

1 

90 
1 

2150.42  4-  32  =  67.2  cubic  inches,  while  the  liquid  quart  con- 
tains 231  4-  4  =  57f  cubic  inches.  A  box  14  inches  square  and 
11.7  (say  llf)  JeeP.  inside  measurement,  holds  one  bushel.  The 
bushel  is  used  in  measuring  charcoal  and,  sometimes,  lime  and 
coal.  The  British  bushel  is  equal  to  8  British  imperial  gallons; 
it  Therefore  contains  277.463  X  8  =  2219.704  cu.  in.  and  one 
British  bushel  equals  2219.704  -s-  2150.42  =  1.03222  -  Winchester 
bushels.  A  cylinder  18|  in.  in  diameter  and  8  in.  deep  holds 
exactly  one  bushel. 

TABLE  VII 

ANlil'LAR  MEASURE 

60  seconds  (") =    1  minute ' 

60  minutes =    1  degree ° 

90  degrees =    1  quadrant L 

4  quadrants  (360°) =    1  circle cir. 

min.  (')  sec.  (") 

21,600  =  1,296.000 

5,400         =  324,000 

60  =  3,600 

1  =  60 

193.  Table  VII  is  very  important  in  connection  with  all  prob- 
lems in  which  the  measurement  of  angles  is  necessary  for  their 
solution.  The  circle  is  divided  into  360  equal  parts  called 
degrees;  each  degree  is  divided  into  60  equal  parts  called  minutes; 
and  each  minute  is  divided  into  60  equal  parts  called  seconds. 
When  the  circle  is  divided  into  four  equal  parts,  each  part  is 
called  a  quadrant,  and  the  4  equal  angles  so  formed  are  called  right 
angles.     A  quadrant  and  a  right  angle  contain  360°  4-  4  =  90°. 

194.  Units  of  Measurement. — It  should  be  evident  from  Art. 
1  that  before  any  measurement  can  be  made  it  is  necessary  to 

establish  a  unit.  For  linear  measurements 
(Table  I),  the  fundamental  unit,  from  which 
the  other  units  are  derived,  is  the  yard;  the 
yard  is  divided  into  three  equal  parts  each  of 
which  is  called  one  foot;  the  foot  is  divided 
into  twelve  equal  parts,  each  of  which  is 
called  one  inch.  The  higher  units,  rods, 
furlongs,  and  miles,  are  obtained  by  taking  a 
certain  number  of  yards. 
The  units  for  square  and  cubic  measure  are  obtained  from  those 

for  linear  measure;  thus,  the  square  inch  is  a  square,  every  side  of 

which  measures  1  inch  (see  Fig.  2). 


Fig. 


§1 


COMPOUND  NUMBERS 


111 


m  p 

c 

■ 

n  g 


Fig.  3. 


The  square  foot  is  a  square  every  edge  of  which  measures  1 
foot  or  12  inches.  If  a  Bquare  foot  be  divided  into  12  equal  parts 
by  lines  ab,  cd,  etc.,  Fig.  3,  and  again  divided  into  12  equal  parts 
by  lines  mn,  pq,  etc.,  every  one  of  the  little  squares  that  are  thus 
formed  will  be  a  square  inch.  Between  a  and  b,  there  are  12 
inch-squares,  between  c  and  d,  12  inch-squares,  etc.  Conse- 
quently, the  foot-square  has  been  divided 
into  12  X  12  =  144  inch-squares,  and 
there  are  144  square  inches  in  1  square 
foot.  A  square  yard  may  be  similarly 
divided  into  9  foot-squares;  hence,  there 
are  9  square  feet  in  1  square  yard.  Note 
that  12  X  12  -  122  and  3  X  3  =  32;  there- 
fore, since  there  are  5.5  yards  in  a  rod,  a 
square  rod  contains  5.52  =  30.25  =  30^ 
square  yards;  etc.  The  number  of  square 
units  in  any  square  may  be  found  by  squaring  the  length  of  the 
side  of  the  square.  It  is  for  this  reason  that  the  second  power  of 
a  number  is  usually  called  the  square. 

A  cube  measuring  1  inch  on  every  edge  is  called  a  cubic  inch 
(see  Fig.  4).  A  cubic  foot  is  a  cube  measuring  1  foot  on  every 
edge.  Such  a  cube  may  be  divided  into  12  equal  layers,  each  of 
which  measures  12  inches  on  each  side  and  1  inch  high.  Each 
layer  may  therefore  be  divided  into  122  equal  cubes  measuring  1 
inch  on  each  edge;  that  is.  each  layer 
may  be  divided  into  12  X  12  inch-cubes; 
and  since  there  are  12  layers,  a  cube 
measuring  12  inches  (=  1  foot)  on  every 
edge  may  be  divided  into  12  X  12  X 
12  =  123  little  cubes  measuring  1  inch 
on  every  edge.  Similarly,  a  cubic  yard 
is  a  cube  measuring  1  yard  on  every 
edge  and  may  be  divided  into  33  =  27 
small  cubes  measuring  1  foot  on  every 
edge.  Hence,  a  cubic  foot  contains 
128  =  1728  cubic  inches,  and  a  cubic  yard  contains  33  =  27 
cubic  feet  =  363  =  4().r>.">ti  cubic  inches.  The  number  of  cubic 
units  in  any  cube  may  be  found  by  cubing  the  Length  of  one  of 
the  edges.  It  is  for  this  reason  that  the  third  power  of  a 
number  is  called  the  cube. 

The  fundamental  unit  of  weight  is  the  pound  avoirdupois;  the 


Fio.  4. 


112  ARITHMETIC  §1 

other  units  of  weight  are  found  by  dividing  or  multiplying  the 
pound.  The  fundamental  unit  of  liquid  measure  in  the  United 
States  is  the  United  States  or  wine  gallon,  which  contains  231 
cubic  inches.  In  Canada  the  law  allows  the  use  of  the  Imperial 
gallon  only  for  a  commercial  unit.  Since 
the  fundamental  unit  of  dry  measure  is  the 
Win ehester  bushel,  which  contains  2150.42 
cubic  inches,  the  units  of  both  liquid  and 
dry  measures  depend  upon  the  unit  of 
linear  measure. 

The  fundamental  unit  of  angular  mea- 
sure is  the  quadrant  or  the  right  angle. 
The  circle  is  divided  into  2  equal  parts  by 
a  line  drawn  through  the  center  called  a 
diameter,  as  ab,  Fig.  5.  Each  half  is  then  divided  into  2  equal 
parts  by  another  diameter  cd,  thus  dividing  the  entire  circle  into 
4  equal  parts.  Each  of  these  four  equal  parts  is  then  divided 
into  90  equal  parts  to  obtain  degrees. 

TABLE  VIII 

United  States  money 

10  mills  (m.) =    1  cent ct.  or  jf. 

10  cents =    1  dime d. 

10  dimes  (100c.) =    1  dollar dol.  or  $ 

10  dollars =    1  eagle E. 

E.         $  d.  ct.  m. 

1  =  10  =  100  =  1000  =  10,000 

1  =  10  =  100  =  1,000 

1  =   io  =    100 

1  =    10 

Note. — With  the  exception  of  the  mill  and  the  eagle,   the   Canadian  unit  of  money  have 
the  same  names  and  relative  values. 

196.  The  fundamental  unit  of  United  States  money  is  the 
gold  dollar,  which  is  defined  as  25.8  grains  of  gold  nine-tenths 
fine;  that  is,  the  gold  dollar  contains  25.8  X  .9  =  23.22  grains  of 
pure  gold.  It  will  be  noted  that  the  scale  in  the  table  of  United 
States  money  is  10,  the  same  as  in  the  Arabic  system  of  notation; 
it  is  therefore  called  a  decimal  scale,  and  numbers  may  be  expressed 
in  different  units  by  simply  shifting  the  decimal  point.  Thus, 
$12.43  =  1243c,  and  978c  =  $9.78.  Since  there  are  100c  in  $1, 
to  change  dollars  to  cents,  multiply  by  100,  that  is,  shift  the 
decimal  point  two  places  to  the  right;  and  to  change  cents  to 
dollars,  shift  the  decimal  point  two  places  to  the  left. 


§1  COMPOUND  NUMBERS  113 

TABLE  IX 

TIME 

60  seconds  (sec.) =   1  minute min. 

60  minutes =    1  hour hr. 

24  hours =1  day da. 

7  days =   1  week wk. 

365  days =    1  year yr. 

yr.       da.  hr.  min.  sec. 

1    =  365    =  8760   =  525,600   =   31,536,000 

1    =       24    =        1,440    =  86,400 

1    =  60    =  3,600 

1    =  60 

197.  The  fundamental  unit  of  time,  universally  used,  is  the 
second.  The  year  does  not  contain  exactly  365  days,  but 
365.242216  days  =  365^  days  very  nearly;  hence,  every  fourth 
year,  one  day  is  added,  making  366  days  in  what  are  called  leap 
years. 

TABLE  X 

MISCELLANEOUS 

12  of  anything =1  dozen doz. 

12  dozen =    1  gross gr. 

12  gross =    1  great  gross g.  gr. 


24  sheets  of  paper =    1  quire qr. 

20  quires  (480  sheets) =1  ream rm. 

198.  A  gross  is  evidently  equal  to  12  X  12  =  144.  It  is  now 
quite  common  practice  to  consider  a  ream  as  500  sheets,  except 
for  stationery  papers. 


THE  METRIC  SYSTEM 

199.  The  metric  system  is  thus  called  because  it  is  based  on  the 
meter  (also  spelled  metre),  which  is  equal  to  39.370113  inches. 
By  Act  of  Congress,  the  meter  contains  39.37  inches.  Sub- 
divisions of  the  meter  are  the  decimeter,  centimeter,  and  millimeter, 
and  multiples  of  it  are  dekameter,  hektometer,  kilometer,  and  myria- 
meter.  The  scale  of  the  system  is  a  decimal  one,  the  same  as  with 
ordinary  numbers  and  United  States  money.  The  prefixes  deci, 
centi,  and  milli  denote  respectively  one-tenth,  one-hundredth, 
and  one-thousandth  of  the  unit  to  which  they  are  prefixed,  the 
same  as  the  dime,  cent,  and  mill  denote-one-tenth,  one-hundredth, 
and  one-thousandth  of  a  dollar.     The  prefixes  deka,  hekto,  kilo, 

8 


114  ARITHMETIC  §1 

and  myria,  denote  respectivly  10,   100,  1000,  and  10,000  times 
the  unit  to  which  they  are  prefixed. 

TABLE  XI 

LINEAR  MEASURE 

10  millimeters  (mm.) =1  centimeter em. 

10  centimeters =1  decimeter dm. 

10  decimeters =1  meter m. 

10  meters =1  dekameter Dm. 

10  dekameters =1  hektometer Hm. 

10  hektometers =    1  kilometer Km. 

10  kilometers =1  myriameter Mm. 

200.  The  fundamental  unit  in  the  foregoing  table  is  the  meter; 
the  other  units  that  are  principally  used  are  the  millimeter  and 
centimeter  for  short  lengths  and  the  kilometer  for  long  dis- 
tances. The  following  table  gives  English  equivalents  for  these 
units. 

1  millimeter    =  .03937  in.   =  jh  in-  very  nearly      =  .04  inch  roughly. 
1  centimeter  =  .3937  in.     =  ff  in.  very  nearly        =  .4  in.  roughly. 
1  motor  =  39.37  in.     =  39|  in.  very  nearly      =  40  in.  roughly. 

1  kilometer     =  39,370  in.   =  H  mi-  very  nearly       =  f  mile  roughly. 
1  meter  =  3.281  ft.  nearly  =  1.094  yd.  nearly. 

It  will  be  observed  that  the  abbreviations  for  the  multiples  of 
the  principal  unit  begin  with  a  capital  letter,  while  those  for  the 
sub-multiples  and  for  the  principal  unit  itself  begin  with  a  lower 
case  letter. 

TABLE  XII 

SQUARE  MEASURE 

100  square  millimeters  (mm2).. .  .    =    1  square  centimeter cm.2 

100  square  centimeters =1  square  decimeter dm.2 

100  square  decimeters =1  square  meter m? 

201.  This  table  does  not  include  measures  of  land,  the  principal 
unit  of  which  is  the  Are  =  100  m.2  The  hectare,  which  equals 
100  ares,  or  100  X  100  =  10,000  square  meters,  takes  the  place 
of  the  English  acre,  and  is  equivalent  to  2.471  acres,  say  2\ 
acres,  roughly.  1  square  meter  =  1550  square  inches,  very 
nearly   =    10.7041  square  feet,  say  lOf  square  feet,  roughly. 

TABLE  XIII 

CUBIC  MEASURE 

1000  cubic  millimeters  (mm.3)....    =    1  cubic  centimeter...   c.c.  or  cm.3 

1000  cubic  centimeters =    1  cubic  decimeter.  .  .   dm.3 

1000  cubic  decimeters =    1  cubic  meter m.3 


1  cubic  motor  =  35.3150  cubic  feet  =  1.308  cubic  vards. 


§1  COMPOUND  NUMBERS  115 

TABLE  XIV 

LIQUID  MEASURE 

10  milliliters  (ml.) =    1  centiliter cl. 

10  centiliters =    1  deciliter dl. 

10  deciliters =1  liter 1- 

10  liters -    1  dekaliter Dl. 

10  dekaliters =1  hektoliter HI. 

10  hektoliters =   1  kiloliter KL 

202.  The  principal  units  of  liquid  measure  are  the  liter  and 
the  hectoliter.  The  liter  is  equal  in  volume  to  1  cubic  decimeter 
=  61.0254  cubic  inches  =  1.0567  quarts  =  l*Hnr  quarts.  The 
hectoliter  =  26.4179  gallons  =  .9435  barrel.  The  milliliter  is 
used  by  physicians,  chemists,  scientists,  and  others  for  measur- 
ing small  amounts.  Since  a  milliliter  is  Wt?  of  a  liter,  and  since 
a  liter  equals  a  cubic  decimeter  =  1000  cubic  centimeters, 
a  milliliter  =  1  cubic  centimeter,  and  it  is  customary  to  call 
this  unit  a  cubic  centimeter  instead  of  a  milliliter.  A  teaspoon 
contains  %  of  a  fluid  ounce.  1  liter  =  1  dm3=  61.0254  cu.  in.; 
1  quart  =  57.75  cu.  in.  =  16  X  2  =  32  fluid  ounces;  %  of 
a  fluid  ounce  =  57.75  -^  32  ^  8  =  .225586  cu.  in.;  1  milliliter  = 
1  cubic  centimeter  =  61.0254  -5-  1000  =  .0610254  cu.  in. 
Therefore,  1  teaspoon  =  .225586  ■*■  .0610254  =  3.6966,  say 
3.7  cubic  centimeters,  and  1  c.c.  =  .27052  teaspoon,  or  a  little 
more  than  H  teaspoon.  Also,  1  fluid  ounce  =  3.6966  X  8 
=  29.5728  cubic  centimeters  =  30  c.c,  nearly. 

TABLE  XV 

MEASURES    OF   WEIGHTS 

10  milligrams  (mg.) =   1  centigram eg. 

10  centigrams .    =1  decigram dg. 

10  decigrams =1  gram g- 

10  grams =    1  dekagram Dg. 

10  dekagrams =1  hektogram Hg- 

10  hektograms =    1  kilogram Kr. 

1000  kilograms =1  tonne T. 

204.  The  fundamental  unit  of  weight  in  the  metric  system 
is  gram,  which  is  the  weight  of  1  cubic  centimeter  of  water, 
weighed  under  certain  conditions,  and  is  equal  to  15.432  grains, 
(more  accurately,  15.432356  gr.).  The  kilogram  (frequently 
called  the  kilo)  =  1000  grams  =  the  weight  of  1  liter  of  water 
=  2.2046  pounds  (more  accurately,  2.20462234  pounds).  The 
tonne  (also  called  metric  ton)  =  1000  kilograms  =  the  weight  of 


116  ARITHMETIC  §1 

1  cubic  meter  of  water  --=  220-1.6  pounds.  The  metric  ton  is 
therefore  approximately  equal  to  the  English  long  ton.  The 
milligram  is  used  by  chemists  for  weighing  minute  quantities; 
it  is  equal  to  nV  of  a  grain  approximately,  or  e,}rg  gr.  very 
accurately. 


REDUCTION  OF  COMPOUND  NUMBERS 

205.  To  reduce  a  compound  number  is  to  change  it  so  that  the 
denominations  used  to  express  it  will  be  different  without  chang- 
ing the  value  of  the  number.  For  example,  it  may  be  desirable 
to  express  a  certain  number  of  miles,  rods  and  feet  as  feet  or 
inches;  or  it  may  be  desired  to  express  the  number  in  miles  and 
fraction  of  a  mile.  All  such  changes  are  called  reduction  or 
reducing  the  number  to  lower  or  higher  denominations. 

206.  Reducing  to  Lower  Denominations. — The  process  is  best 
illustrated  by  means  of  an  example.     Thus,  express  3  mi.  126  rd. 

9   ft.    in   feet.     Arranging   the   work    as 

shown,  multiply  the  number  of  rods  in  a 

mi.       rd.        ft.  mile  (320)  by  the  number  of  miles  in  the 

3        126        9  given   number,   in  this  case  3,   and  the 

product  is  the  number  of  rods  in  3  miles. 

'Mi0rd-  To  this  add  the  number  of  rods  in  the 

given  number,  and  the  sum,  1086,  is  the 

jjw  '  number  of  rods  in  3  mi.  126  rd.     Since 

-,..  "  there  are  16^  feet  in  a  rod,  the  product 

6516  of  1086  and  16J£  =  17919  is  the  number 

1086  of  feet  in  1086  rd.     To  this  add  the  9  ft. 

17919  ft.  in    the    given    number,    and    the  sum  is 

■' ft-  17,928  ft.,  the  number  of  feet  in  3  mi.  236 

17928  ft.    Ana.  n\.  9  ft.     Had  it  been  desired  to  reduce 

the    number   to    inches,    the   last    result 

would  have  been  multiplied  by  12,  and 

as  there  are  do  inches  in  the  given  number,  the  final  result  would 

have  been  17928  X  12  =  215,136  in.  =  3  mi.  126  rd.  9  ft. 

207.  Rule. — To  reduce  a  compound  number  to  lower  denomina- 
tions, find  the  product  of  the  number  representing  the  highest  denomr 
ination  in  the  given  number  and  the  number  of  units  of  the  next 
lower  denomination  that  will  make  one  unit  of  the  next  higher  denom- 
ination, and  add  to  this  product  the  number  of  units  of  that  lower 
deno?nination,   (if  any)  in  the  given  number.     Repeat  this  process 


00 


§1  COMPOUND  NUMBERS  117 

to  reduce  to  the  next  lower  denomination,  continuing  in  this  manner 
until  the  required  denomination  has  been  used. 
Example  1. — Reduce  32°  47'  19"  to  seconds. 

f-  Solution. — The  work  is  shown  in  the  margin,  and  should 

be  evident.  Since  there  are  GO  minutes  in  1  degree,  the 
product  of  60  and  32  is  first  found,  to  which  is  added  the  47.' 
Since  there  are  60"  in  1  minute,  the  product  of  60  and 
1967  is  then  found,  to  which  is  added  the  19".  The  num- 
ber is  now  reduced  to  the  denomination  required.  Note 
that  like  numbers  are  added — minutes  to  minutes  and 
seconds  to  seconds.  It  is  not  customary  to  write  the 
1 1  SOW     ,4ns      abbreviations  when  reducing. 

Example  2. — Express  156  m.,  5  cm.,  2  mm.  in  millimeters. 

Solution. — Write  the  number  without  abbreviations,  with  a  period  fol- 
lowing the  number  of  units  of  the  highest  denomination,  and  with  ciphers 
in  place  of  any  denomination  that  is  missing.  Since  there  are  no  decimeters, 
156  m.  5  cm.  2  mm.  =  156.052  meters.  The  process  is  exactly  the  same  as 
writing  a  number  in  the  Arabic  system  of  notation.  To  reduce  156.052  m. 
to  millimeters,  multiply  by  1000  by  shifting  the  decimal  point  3  places  to 
the  right,  and  156.052  m.  =  156.052  mm.     Ans. 

Had  it  been  desired  to  express  the  above  number  in  decimeters, 

multiply  it  by  10  by  moving  the  decimal  point  1  place  to  the 

right,    obtaining    1560.52    dm.;   to   express   it   in   centimeters, 

move   the   decimal  point  two  places  to  the  right,   obtaining 

15605.2  cm. 

208.  Reducing  to  Higher  Denominations. — The  process  is  best 

illustrated  by  an  example.     Thus,  reduce  215,136  in.  to  higher 

denominations. 

Since   there   are    12  inches   in    1 

215136  in.  (12  foot,  the  number  of  feet  in  215,136 

1792S.0  ft.  (16.5  inches    may  be  found  by  dividing 

l6^  1086  215,136  by  12;  there  is  no  remainder, 

1428  and  the  quotient  is  17,928  ft.     There 

1320 

are   16.5  feet  in  1  rod;  hence,  the 

^  number   of   rods   in    17,928   feet  is 

90  =  9.0  ft.  found  by  dividing  17>928  hy  16-5- 

10S6  rd.  (320  The  quotient  is  1086  rd.,  and  the 

9°0       3^nL  remainder  is  apparently  90  ft.     In 

126  rd.  reality,  however,  it  is  9  ft.,  because 

3  mi.  126  rd.  9  ft.    Ans.  the  cipher  in  90  is  the  one  following 

the  decimal  point  in  the  dividend, 
and  has  no  value.     Since  there  are  320  rods  in  1  mile,  divide 


118  ARITHMETIC  §1 

1086  rods  by  320  to  find  how  many  miles  there  are  in  1086  rd.; 
the  quotient  is  3  mi.,  and  the  remainder  is  126  rd.  Therefore, 
215,136  in.  =  3  mi.  126  rd.  9  ft. 

Note  that  after  any  division  has  been  completed,  the  remainder 
is  of  the  same  denomination  as  the  dividend;  this  is  necessarily 
the  case,  since  the  remainder  is  a  part  of  the  dividend.  The 
quotient,  however,  is  of  a  higher  denomination  than  the 
dividend. 

209.  Suppose  that  instead  of  reducing  the  inches  to  a  compound 
number  the  highest  denomination  of  which,  in  this  case,  is  a  mile, 
it  had  been  desired  to  express  the  number  in  miles  and  a  fraction 
(decimal)  of  a  mile.  In  such  case,  proceed  in  exactly  the  same 
manner  as  before,  except  that  the  quotient  will  in  all  cases  be  a 
mixed  number  (when  there  is  a  remainder),  the  division  being 
carried  to  as  many  decimal  places  as  is  desired.  Thus,  in  the 
example  just  given,  9  ft.  =  9  -*-  16.5  =  .545454+  rd.  Adding 
this  to  the  126  rd.,  the  sum  is  126.545454+  rd.  Then,  126.545454 
-s-  320  =  .39545454+  mile,  which  added  to  the  3  mi.  makes 
3.3954545  mi. 

210.  To  reduce  a  decimal  of  one  denomination  to  units  of  a 
lower  denomination,  proceed  exactly  in  accordance  with  the  rule 
of  Art.  207.  Thus,  .3954545  mi.  =  .3954545  X  320  =  126.54544 
rd.;  .54544  X  16.5  =  8.99976,  say  9  ft.  The  result  would  have 
been  exactly  9  ft.  if  instead  of  expressing  the  rods  and  feet  as  a 
decimal  of  a  mile  they  are  reduced  to  a  fraction  of  a  mile.     Thus, 

9  +  16'5  =  I6T5  =  **  =  "  rd-  126  +  A  =  126"  =  ****  rd- 
1  ?  | "  •*-  320  =  ifH  =  -&V  =  -39A  mi.  This  last  expression  will 
reduce  to  126  rd.  9  ft.  exactly. 

211.  Rule  I. — To  reduce  a  denominate  number  to  higher  denomi- 
nations, begin  with  the  lowest  denomination  of  the  given  number  and 
divide  the  number  of  units  of  that  denomination  by  the  number  of 
units  required  to  make  one  unit  of  the  next  higher  denomination, 
forming  either  a  common  fraction  {which  reduce  to  its  lowest  terms) 
or  carrying  the  quotient  to  any  desired  number  of  decimal  -places. 
Add  the  quotient  thus  obtained  to  the  number  of  units  in  the  given 
number  of  the  same  denomination  as  the  quotient  (if  any),  and  dinde 
the  sum  by  the  number  of  units  required  to  make  one  unit  of  the  next 
higher  denomination.  Proceed  in  this  manner  until  the  desired 
denomination  is  reached. 


§1  COMPOUND  NUMBERS  119 

II.  //,  however,  it  is  desired  to  reduce  a  given  number  of  units  of  a 
lower  denomination  to  a  compound  number  of  higher  denomination, 
divide  the  given  number  by  the  number  of  units  required  to  make  one 
unit  of  the  next  higher  denomination;  the  quotient  will  be  of  the  next 
higher  denomination,  and  the  remainder  (if  any)  will  be  of  the  same 
denomination  as  the  dividend.  If  the  quotient  is  larger  than  the 
number  of  units  required  to  make  a  unit  of  the  next  higher  denomina- 
tion, divide  it  by  the  number  of  units  required  to  make  one  unit  of  the 
next  higher  denomination.  Proceed  in  this  manner  nutil  the  highest 
denomination  is  reached  or  a  quotient  is  obtained  that  is  smaller 
than  the  number  of  units  required  to  make  a  unit  of  the  next  higher 
denomination. 

Example. — Reduce  123,456"  to  a  compound  number;  also  express  it  in 
degrees  and  decimal  of  a  degree  to  5  decimal  places. 
60)  1 23456"  Solution. — Dividing  first  by  60,  the  quotient 

60)2057'  +  36"  is  205/'  and  the  remainder  is  36".     Dividing 

34°  _}_  17'  again  by  60,  since  there  are  60'  in  1°,  the  quo- 

34°  17' 36".     Ans.  tient  is  34°  and  the  remainder  is  17'.     Since  34° 

60)  123456"  is  leSS  than  90°' the  number  of  degrees  necessary 

'  to    make    a    quadrant,    the   work    ceases,    and 

60)  2057.6'  123,456"  =  34°  17'  36".     The  work  for  reducing 

34.29333°  +  Ans.  to  degrees  and  decimal  of  a  degree  is  evident. 
Example  2. — Express  3250615  milligrams  in  kilograms. 
Solution.— Beginning  with  the  right  hand  figure,  move  the  decimal  point 
one  place  to  the  left  for  each  higher  denomination  until  the  desired  denomi- 
nation is  reached.  Thus,  say  milligrams,  centigrams,  decigrams,  grams, 
dekagrams,  hektograms,  kilograms,  placing  the  pencil  point  on  the  position 
occupied  by  the  decimal  point  in  each  case,  it  being  to  the  right  of  the 
milligrams  in  the  beginning.  When  kilograms  is  reached,  the  pencil  point 
will  fall  between  5  and  3;  consequently,  5320615  mg.  =  5.320615  Kg.  Ans. 
This  same  method  may  be  used  in  reducing  metric  numbers 
(except  those  for  square  and  cubic  measure)  to  lower  denomina- 
tions. For  square  measure,  move  the  decimal  point  two  places 
each  time  the  name  of  a  denomination  is  pronounced,  and  for 
cubic  measure,  move  it  three  places.  For  instance,  59308726 
mm2  =  59.308726  m2,  and  607849358  c.c.  =  607.849358  m3. 
Also,  78.06342  m2  =  780634.2  cm2  =  78063420  mm2,  and 
9.50783  m3  =  9507830  c.c.  =  9507830000  mm3. 

212.  Another  way  of  reducing  the  lower  units  of  a  compound 
number  to  a  decimal  or  fraction  of  higher  denomination  is  the 
following:  To  express  23  rd.  3  yd.  1  ft,  8  in.  as  a  decimal  of  a 
mile,  first  reduce  the  compound  number  to  the  lowest  denomi- 
nation in  the  given   number,   in  this  case  inches,  and  23  rd.  3 


120  ARITHMETIC  §1 

yd.  1  ft.  8  in.  =  4682  in.  Referring 
to  Table  I,  and  consulting  the  sub- 
sidiary table,  it  is  seen  that  there  are 
63360  inches  in  a  mile;  hence,  4682 
in.  is  sVsVirth  of  a  mile.  Reducing 
this  fraction  to  a  decimal,  vsVA  = 
.073895+.  Therefore,  23  rd.  3  yd. 
1  ft,  8  in.  =  .073895  mi. 

This  method  is  to  be  preferred  to 
the  former  one,  when  reducing  to  a 
decimal  or  a  fraction  of  a  higher 
denomination,  since  it  entails  no 
more  work,  if  as  much,  and  is,  in 
general,  more  accurate. 


rd.  yd. 

ft. 

23   3 

1 

5.5 

115 

115 
126.5 

3 

129.5  yd. 
3 

388.5 

1 

389.5  ft. 

12 

4674 . 0 

8 

4082   in. 

EXAMPLES 


(1)  Reduce  5  mi.  3  fur.  36  rd.  4  yd.  2  ft.  7  in.  to  inches.     Ans.  347,863  in. 

(2)  Reduce  the  number  in  (1)  to  miles  and  decimal  of  a  mile. 

Ans.  5.490276  mi. 

(3)  Reduce  13  bbl.  23  gal.  1  pt.  to  pints.  Ans.  3,461  pt. 

(4)  Reduce  the  number  in  (3)  to  barrels  and  decimal  of  a  barrel. 

Ans.  13.73413  bbl. 

(5)  Reduce  3  T.  13  cwt.  71  lb.  12  oz.  to  ounces.  Ans.   117,948  oz. 

(6)  Reduce  the  number  in  (5)  to  tons  and  decimal  of  a  ton. 

Ans.  3.685875    T. 

(7)  Reduce  14°  9'  54"  to  seconds.  Ana.  50,994". 

(8)  Express  14°  9'  54"  in  degrees  and  decimal  of  degree.      Ans.   14.165°. 

(9)  Reduce  75,906"  to  a  compound  number.  Ans.  21°  5'  6". 

(10)  Reduce  75,906"  to  degrees.  Ans.  21.0S5°. 

(11)  Reduce  50,000  in.  to  a  compound  number. 

Ans.  6  fur.  12  rd.  2  yd.  2  ft.  8  in. 


OPERATIONS  WITH  COMPOUND  NUMBERS 

213.  Addition. — Compound  numbers  are  added  in  practically 
the  same  manner  as  abstract  numbers.  The  numbers  to  be 
added  are  arranged  under  one  another  with  like  denominations  in 
the  same  columns.  Then,  beginning  with  the  lowest  denomina- 
tion, add  the  numbers  in  that  column;  if  the  sum  is  greater  than 
the  number  of  units  required  to  make  a  unit  of  the  next  higher 
denomination,  reduce  it  to  the  next  higher  denomination  and  carry 
to  the  next  column  the  number  of  units  so  found  of  that  next 


§1  COMPOUND  NUMBERS  121 

higher  denomination.     The  process  is  repeated  for  the  second  and 
subsequent  columns. 

Example  1.— Add  15°  20'  36",  21°  53'  46",  8°  49'  28",  and  76°  51'  17". 

Solution*. — The  numbers  are  arranged  as  shown  in  the  margin,  with 

15°  20'  36"  seconds  under  seconds,  minutes  under  minutes,  etc. 

21     53    46  ^e  sum  °^  *ne  second's.  column  is   127"  =  2'  7"; 

o     .q    OS  write  the  7"  and  carry  the  2'  to  the  next  column, 

«g     r-j     17  adding  the  2'  to  the  numbers  in  that  column.     The 

,,.,  ",'  sum  thus  obtained  is  175'  =  2°  55'.     The  2°  is  carried 

to  the  next  column,  making  the  sum  122°,  and  the  sum 

of  all  the  numbers  is  122°  55'  7".     It  is  not  customary 

to  write  the  sums  of  the  columns,  as  shown  here;  only 

the   final  results   are   written,   unless  there  are  fractions   as  in  the  next 

example. 

Example  2.— Add  11  rd.  4  yd.  2  ft.  4  in.,  34  rd.  1  yd.  9  in.,  27  rd.  3  yd.  1  ft. 
6  in.,  and  17  rd.  1  yd.  1  ft.  2  in. 

Solution. — The  numbers  arc  arranged  as  in  example  1,  and  the  sum  as 

found  is  2  fur.  10  rd.  4.5  yd.  2  ft.  9 
in.  Since  it  is  inconvenient  to  have 
a  fraction  in  any  but  the  lowest  de- 
nomination, reduce  the  .5  yd.  obtain- 
ing 1  ft.  6  in.,  which  is  added  as 
shown,  making  the  final  sum  2  fur. 
10  rd.  5  yd.  1  ft.  3  in. 

Ans. 

Rule. — Place  the  numbers  to  be  added  under  each  other,  with  like 
denominations  in  the  same  columns.  Add  each  column,  beginning 
with  that  of  the  lowest  denomination.  If  the  sum  of  the  numbers  in 
any  column  is  greater  than  the  number  of  units  required  to  make  a 
unit  of  the  next  higher  denomination,  reduce  the  sum  to  the  next 
higher  denomination  before  adding  the  next  column.  When  the 
sum  has  been  found,  if  the  number  of  units  in  any  denomination 
contains  a  fraction,  reduce  the  fraction  of  a  unit  to  loiver  denomina- 
tions and  add  to  the  units  of  lower  denomination  in  the  sum  previously 
found. 

214.  Subtraction. — The  operation  of  subtraction  is  practically 
the  same  as  in  subtraction  of  abstract  numbers.  Place  the  sub- 
trahend under  the  minuend,  with  like  denominations  under  each 
other.  If  the  number  of  units  of  any  denomination  in  the  sub- 
trahend is  larger  than  the  number  above  it,  reduce  one  unit  of 
the  next  higher  denomination  to  the  next  lower,  add  it  to  the 
number  of  units  of  that  denomination  in  the  minuend,  and  then 


rd. 

yd. 

ft.  in. 

11 

4 

2  4 

34 

1 

9 

27 

3 

1  6 

17 

1 

1   2 

90 

10 

5  21 

2  fur.  10 

4$ 

2   9 

5yd. 

=  1  6 

2  fur.  10 

rd 

•  5  yd. 

1  ft.  3  in. 

rd. 
34 
27 

yd.    ft. 
1         0 
3         1 

in. 
9 
6 

6 

2K     2 
1 

3 
6 

6rd. 

3  yd. 

9 

122  ARITHMETIC  §1 

subtract.  Add  1  to  the  number  of  units  in  the  next  column  of 
the  subtrahend  before  subtracting.  This  process  is  exactly  similar 
to  that  employed  in  subtracting  abstract  numbers. 

Example  1.— Subtract  34°  27'  17"  from  90°. 

Solution. — Placing  the  subtrahend  under  the  minuend,  there  are  no 
seconds  in  the  minuend;  hence,   1'  =  60"  is  added  to 
90°  00'  00"  the  minuend,  and  60"  -  17"  =  43".     Adding  1'  to  27', 

34    27    17  the  sum   is  28',   and   60'  -  28'  =  32'.     Adding    1°  to 

55°  32'  43".     Ans.  34°,   the  sum  is  35°,   and   90°  -  35°  =  55°.     The  re- 
mainder, therefore,  is  55°  32'  43". 

Example  2. — From  34  rd.  1  yd.  9  in.  subtract  27  rd.  3  yd.  1  ft.  6  in. 
Solution. — Arranging  the  numbers  as  in  example  1,  1  ft.  cannot  be  sub- 
tracted from  0  ft.;  hence,   1  yd.  =  3  ft.  is 
added  to  the  minuend,  and  3  ft.  —  1  ft.  =  2 
ft.     Adding  1  to  the  3  yd.  makes  4  yd.;  but 
4  yd.  cannot  be  subtracted  from  1  yd.;  hence, 
1   rd.  =  5H  yd-   is  added  to  the  minuend, 
making  1  +  5%  =  6^2  yd.,  and  6}i  yd.  — 
Ans.     4  yd.  =  2%  yd.     Finally,  27  rd.  +  1  rd.  = 
28  rd.,   and   34   rd.  -  28   rd.  =  6   rd.     The 
L2  yd.  is  reduced  to  1  ft.  6  in.  and  added,  the  sum  being  3  ft.  9  in.     But  3  ft. 
=  1  yd.;  hence,  the  number  of  yards  is  increased  by  1,  making  the   final 
sum  6  rd.  3  yd.  9  in. 

Rule. — Place  the  subtrahend  under  the  minuend,  with  like 
denominations  in  the  same  columns.  Beginning  with  the  column 
of  lowest  denomination,  subtract  the  numbers  in  the  bottom  row  from 
those  above  them.  If  the  number  in  any  column  of  the  subtrahend 
is  larger  than  the  number  above  it  in  the  minuend,  reduce  one  unit 
of  the  next  higher  denomination  to  the  next  lower  denomination, 
add  it  to  the  minuend,  then  subtract,  and  add  1  to  the  number  in  the 
column  of  the  next  higher  denomination  in  the  subtrahend.  Con- 
tinue in  this  manner  until  the  entire  remainder  has  been  found. 

215.  Multiplication. — A  compound  number  may  be  multiplied 
by  an  abstract  number  in  two  ways:  1st.,  by  multiplying  the 
units  of  each  denomination  separately,  and  reducing  the  products 
to  higher  denominations;  this  is  advisable  when  the  multiplier  is 
a  small  number.  2d,  reduce  the  compound  number  so  that  it 
will  be  expressed  in  units  of  one  denomination,  preferably,  the 
lowest  denomination  in  the  compound  number,  then  multiply, 
and  reduce  the  product  to  higher  denominations;  this  process 
is  preferred  when  the  multiplier  is  a  large  number,  say  greater 
than  12,  or  when  it  contains  a  fraction  or  a  decimal. 


§1  COMPOUND  NUMBERS  123 

Example  1.— Multiply  56  T.  13  cwt.  72  lb.  by  8. 

Solution.— The  product  of  8  and  72  lb.  is  576  lb.  =  5  cwt.  76  lb.     Then 

T.     cwt.     lb.  13  cwt-  X  8  =  104  cwt.,  to  which  is  added  the 

56       13       72  5  cw*-  carried  from  the  first  product,  making 

8  109  cwt.  =  5   T.   9  cwt.     Lastly,   56  T.  X  8 

-^     —     r_-T"  =  448  T.,  to  which  is  added  the  5  T.  carried 

453  T.  9  cwt.  76  lb.     Am.     5?m,J ^   Preced,ing    P™d™t,    making    453 

T.     The  final  product  is  453  T.  9  cwt.  76  lb. 

Example  2.— Multiply  7  gal.  3  qt.  1  pt.  by  53. 

Solution.— Reducing  to  pints,  7  gal.  3  qt.  1  pt.  =  63  pt.;  63  pt.X53 
=  3339  pt.  =  6  hhd.  1  bbl.  7  gal.  3  qt.  1  pt.  When  reducing  3339  pt.  to 
higher  denominations,  the  first  result  obtained  is  6  hhd.  1  bbl.  iy2  gal.  1  qt. 
1  pt.  Since  y2  gal.  =  2  qt.,  the  final  result  without  fractions  is  6  hhd.  1  bbl. 
7  gal.  3  qt.  1  pt.     Ans. 

Rule  I. — Begin  with  the  lowest  denomination  in  the  given  num- 
ber, and  multi-ply  the  number  of  units  of  that  denomination  by  the 
multiplier;  reduce  the  product  to  the  next  higher  denomination  by 
dividing  by  the  number  of  units  required  to  make  1  of  the  next 
higher  denomination,  and  add  the  quotient  to  the  product  of  the 
number  of  units  in  the  next  higher  denomination  and  the  multiplier. 
Proceed  in  this  manner  until  the  complete  product  has  been  found. 
If  a  fraction  occurs  in  connection  with  any  product,  reduce  it  to 
lower  terms. 

II.  If  the  muliipliei'  is  greater  than  12,  or  if  it  contains  a  fraction 
or  a  decimal,  reduce  the  multiplicand  to  the  lowest  denomination 
given,  multiply,  and  reduce  the  product  to  higher  denominations. 

Example.— Multiply  5  yd.  2  ft.  9  in.  by  11.7. 

Solution. — Reducing  to  inches,  5  yd.  2  ft.  9  in.  =  213  in.;  213  in.  X 
11.7  =  2492.1  in.  =  69  yd.  8.1  in.,  or  12  rd.  3  yd.  8.1  in.     Ans. 

216.  Division. — There  are  two  cases  of  division:  dividing  a 
compound  number  by  an  abstract  number,  in  which  case,  the 
quotient  is  a  compound  number;  or  dividing  a  compound  number 
by  a  compound  number  of  the  same  kind,  in  which  case,  the 
quotient  is  an  abstract  number.  The  simplest  method  of  per- 
forming the  division  (and  in  most  cases,  the  easiest)  is  to  reduce 
the  dividend  to  the  lowest  denomination  in  the  given  number, 
divide,  and  reduce  the  quotient  to  higher  denominations.  If 
the  divisor  is  also  compound,  reduce  it  to  the  same  denomination 
as  the  dividend,  and  divide  as  in  division  of  abstract  numbers,  the 
quotient  being  abstract. 

Example  1.— What  is  fth  of  143°  25'  41"? 

Solution.— 143°  25'  41"  =  516,341";  516,341"  -s-  7  =  73,763"  =  20° 
29'  23".     A?is.     A  somewhat  easier  method  of  performing  this  division 


124  ARITHMETIC  §1 

(.which  may  be  used  whenever  the  divisor  is  small)  is  shown  in  the  margin. 
Here  proceed  as  in  short  division.     Then,   143°  -r-  7  =  20°  +  3°  remain- 
] '  ,',  der;  3°  =  180',  which  added  to  the  25',  makes  205'. 

7)143°  25'  41"  205'  -5-  7  =  29'  +  2'   remainder;    2'  =  120",    which 

20°  29'  23".     Ans.  added  to  the  41",  makes   161";  and   161"  4-  7  = 
23".     The  w'ork  may  be  arranged  as  shown  or  the 
additions  may  be  performed  mentally.     The  latter  practice  is  not  recom- 
mended, however,  as  it  tends  to  increase  the  liability  of  making  a  mistake. 
Example  2. — How  many  flasks  holding  2  gal.  1  qt.  1  pt.  may  be  filled  from 
a  cask  holding  52  gal.  1  qt.? 

Solution. — Reducing  both  numbers  to  pints,  2  gal.  1  qt.  1  pt.  =  19  pt.; 
52  gal.  1  qt.  =  418  pt. ;  418  -^  19  =  22;  hence,  22  flasks  may  be  filled.     Ans. 

Rule. — If  the  divisor  is  abstract,  reduce  the  dividend  to  the  lowest 
denomination  in  the  given  number,  divide,  and  reduce  the  quotient 
to  higher  denominations.  If  both  dividend  and  divisor  are  com- 
pound, reduce  them  to  the  lowest  denomination  in  either  number  and 
divide;  the  quotient  will  be  abstract. 


EXAMPLES 

(1)  Find  the  sum  of  17  cwt.  26  lb.  9  oz.,  15  cwt.  83  lb.  11  oz.,  22  cwt.  55  lb. 
6  oz.,  24  cwt.  7 1  lb.  8  oz.,  and  18  cwt.  48  lb.  5  oz.  Ans.  4  T.  18  cwt.  85  lb.  7  oz. 

(2)  What  is  &th  of  360°?  Ans.  32°  43'  38tV. 

(3)  Add  19  rd.  4  yd.  1  ft.  4  in.,  31  rd.  2  yd.  9  in.,  26  rd.  5  yd.  2  ft.  6  in., 
and  35  rd.  1  yd.  1  ft.  10  in.  Ans.  2  fur.  33  rd.  3  yd.  5  in. 

(4)  The  sum  of  the  three  angles  of  any  plane  triangle  is  180°;  if  two  of 
the  angles  of  a  triangle  are  36°  13'  26"  and  64°  45'  40",  what  is  the  other 
angle?  Ans.  79°  0'  54". 

(5)  A  piece  of  tape  line  2  rd.  1  yd.  2  ft.  5  in.  long  was  used  to  measure  the 
distance  between  two  points  about  one  mile  apart.  The  tape  was  applied 
172  times;  what  was  the  distance  between  the  points? 

Ans.   1  mi.  2  fur.  2  yd.  1  ft.  8  in. 

(6)  The  distances  between  the  centers  of  the  faces  of  a  number  of  pulleys 
on  a  main  shaft  were  as  follows:  5  ft.  8£  in.,  8  ft.  6|  in.,  7  ft.  3£  in.,  10  ft. 
9J  in.,  9  ft.  4J  in.,  6  ft.  7\  in.  What  was  the  distance  between  the  first  and 
last  pulleys?  Ans.  48  ft.  3f  in. 

(7)  A  coal  hod  was  found  to  hold  38  lb.  13  oz.  of  coal;  how  many  times 
will  a  ton  of  coal  fill  the  hod?  Ans.  51.5+times. 

(8)  Bought  a  car  of  pulp  weighing  47,526  lb.  on  which  the  freight  rate  was 
1 7  emits  per  hundredweight;  how  much  was  the  bill  for  freight?  Remember 
thai  all  freight  rates  are  computed  on  the  basis  of  2240  pounds  to  the  ton, 
hence,  a  hundredweight  here  contains  112  pounds.  Ans.  $72.14. 

(9)  The  weight  of  a  cargo  of  pulp  was  3923  tons  1458  pounds;  what  was 
the  freight  bill,  if  the  rate  was  $12.75  per  ton?  Ans.  $50,026.55. 

(10)  What  was  the  cost  of  a  belt  95  ft.  7lA  in.  long  at  32**.  per  foot? 

Ans.  $30.60. 

(11)  A  machine  makes  1836  pounds  of  paper  per  hour;  in  what  time  will 
it  make  27  '-i  tons  of  2240  pounds.  Ans.  33  hr.  33  min. 


§1  THE  ARITHMETICAL  MEAN  125 

THE  ARITHMETICAL  MEAN 

217.  The  arithmetical  mean  of  several  numbers  (or  quantities) 
is  the  quotient  obtained  by  dividing  the  sum  of  the  numbers  (or 
quantities)  by  the  number  of  numbers  (or  quantities);  it  is  the 
same  as  the  average  of  several  numbers  (or  quantities),  and  is 
usually  called  the  mean.  For  example,  if  goods  were  bought  from 
three  different  firms  at  discounts  of  32%,  35%,  and  36>£%,  the 
mean,  or  average,  discount  is  (32  +  35  +  36.5)  +  3  =  34.5%. 
This  result  presupposes  that  the  same  amounts  were  bought  from 
each  firm. 

Now  suppose  that  from  one  firm,  the  bill  was  $126.40  and  the 
discount  was  32%;  from  another  firm,  the  bill  was  $87.65  and  the 
discount  was  35%:  from  the  third  firm,  and  bill  was  $68.83  and 
the  discount  was  36.5%;  what  was  the  mean  discount?  Here 
it  is  necessary  to  multiply  the  amount  of  each  bill  by  the  discount 

(a) 

126.40  X  32%  =  4044.8%  $126.40  X  .68     =  $  85.95 

87.65X35       =3067.75  87.65  X  .65     =      56.97 

68.83  X  36.5  =  2512.295  68.83  X  .635  =      43.71 


282.88  X  x        =  9624.845  $186.63 

9624.845  -r-  282.88  =  34.024%  =  x 
$282.88  X  (1  -  .34024)  =  $186.63. 

and  then  divide  the  sum  of  the  products  by  the  sum  of  the  bills. 
The  result  as  shown  herewith  is  a  mean  discount  of  34.024%. 
To  prove  that  this  is  correct,  multiply  the  amount  of  each  bill 
by  1  minus  the  discount  and  add  the  products;  the  sum  will 
evidently  be  the  total  amount  to  be  paid,  and  it  must  equal  the 
total  amount  less  the  mean  discount,  which  is  shown  to  be  true, 
both  being  equal  to  $186.63.  The  process  is  exactly  the  same 
as  would  be  followed  in  finding  the  average  (mean)  price  paid  for 
126.4  lb.  of  some  article  at  32  cents  per  pound,  87.65  lb.  at  35 
cents  per  pound,  and  68.83  lb.  at  36.5  cents  per  pound.  The 
reason  for  the  process  is  easily  found.  If  the  discount  on  $1  is 
32%,  the  discount  on  $126.40  is  126.4  X  32%  =  4044.8%. 
The  total  discount  on  $282.88  is  9624.845%;  hence,  the  average, 
or  mean,  discount  is  9624.845%  -f-  282.88  =  34.024%  on  $1. 

Example  1. — A  machinist  receives  52fl  per  hour  in  wages.  For  every 
hour  he  works  over  8  hours  a  day,  he  receives  time-and-a-half.  His  over- 
time for  one  week  was:  Monday,  2%  hr.;  Tuesday,  2  hr.;  Wednesday,  3  hr.; 
Thursday,  2  hr.;  Friday,  \\i  hr.;  Saturday,  1  hr.  What  was  his  average 
rate  per  hour  for  that  week? 


126  ARITHMETIC  §1 

Solution. — The  total  number  of  regular  hours  of  work  was  6  X  8  =  48, 
for  which  he  received  48  X  52(1  =  $24.96.  The  number  of  hours  that  he 
worked  overtime  was  2.5+2+3+2  +  1.5  +  1  =  12;  his  rate  for  overtime 
was  1.5  X52(f  =  78fi;  and  he  received  for  overtime  work  12  X  78(f  =  $9.36. 
The  total  amount  that  he  received  for  the  week  was  $24.96  +  $9.36  = 
$34.32.  The  total  number  of  hours  worked  was  48  +  12  =  60.  Therefore, 
his  average,  or  mean,  rate  per  hour  for  that  week  was  $34.32  -e-  60  =  $.572 
=  57.2  cents.     A  ru. 

Example  2. — It  was  desired  to  measure  very  accurately  the  distance 
between  two  punch  marks.  The  result  of  measurements  by  five  different 
persons  was  as  follows:  10  ft,  8.1  in.;  10  ft.  8.16  in.;  10  ft.  7.97  in.;  10  ft. 
8.21  in.;  10  ft.  8.05  in.  Which  of  these  measurements  is  nearest  to  the 
mean,  or  average,  of  all  the  measurements? 

Solution". — The  sum  of  all  the  measurements  is  50 
ft.  40. 49  in.  The  number  of  measurements  is  5;  hence, 
the  mean  is  50  ft,  40.49  in.  v5  =  10  ft,  8.098  in.  which 
is  very  nearly  equal  to  the  first  measurement,  and  is 
nearer  that  than  any  of  the  others.  The  probable 
correct  value  is  10  ft.  8.1  in.  Note  that  the  40  in.  was 
not  reduced  to  feet,  because  it  is  to  be  divided  by  5. 

Example  3. — The  floor  area  of  a  room  in  which  36  clerks  are  employed  is 
2960  square  feet;  what  is  the  average  number  of  square  feet  per  clerk? 

Solution. — Evidently,  the  average  number  of  square  feet  per  clerk  is 
2960  -r  36  =  82f ,  or  a  space  about  9  ft.  square  for  each  clerk,  since  92  = 
81.     Ans. 


10  ft. 

8.1      in, 

10 

8.16 

10 

7.97 

10 

8.21 

10 

8.05 

50 

40.49 

10  ft. 

8 .  098  in. 

Ans. 

EXAMPLES 

(1)  The  daily  production  of  a  paper  machine  for  one  week  was  as  follows: 
97.6  T,  101.2  T.,  98.5  T.,  90  T.,  103.1  T,  96.4  T.;  what  was  the  average 
daily  production  for  the  week?  Ans.  97.8  T. 

(2)  The  amount  of  sulphur  used  in  making  sulphite  pulp  was:  In  January, 
149.721  lb.  of  sulphur  for  508  tons  of  pulp;  in  February,  141,176  lb.  for  476 
T;  in  March,  152.148  lb.  for  519  T.;  in  April,  148,635  lb.  for  493  T. ;  in  May, 
147,204  lb.  for  527  T.;  in  June,  153,630  lb.  for  468  T;  in  July,  152,582  lb. 
for  483  T;  in  August,  151,796  lb.  for  479  T;  in  September,  154,881  lb.  for 
492  T;  in  October,  150,300  lb.  for  512  T.;  in  November,  153,714  lb.  for  483 
T. ;  in  December,  149,566  lb.  for  506  T.  What  was  the  average  amount  of 
sulphur  used  per  ton  for  the  year?  Ans.  303.625  —  lb.  per  T. 

(3)  Sold  1250  lb.  of  paper  at  22^  per  pound;  6700  lb.  at  20ff;  10,0001b. 
at  18j5;  5500  lb.  at  21  i\  and  15,000  lb.  at  17£.  What  was  the  average  price 
received  per  pound.  Ans.  18.5176  — ft. 


ARITHMETIC 

(PART  3) 


EXAMINATION  QUESTIONS 

(1)  Referring  to  example  1,  Art.  164,  what  is  the  diameter 
to  the  nearest  64  th  of  an  inch  of  a  circle  whose  area  is  218.7 
sq.  in.?  Ans.  16ff  =  16H  in. 

(2)  Referring  to  example  2,  Art.  164,  with  what  velocity  will 
a  ball  of  lead  strike  the  ground  if  it  fall  from  a  height  of  338  ft.  ? 

Ans.  147.45—  ft.  per  sec. 

(3)  The  area  of  a  circle  is  proportional  to  the  square  of  the 
diameter.  The  area  of  a  circle  whose  diameter  is  29|  in.  is  701 
sq.  in.;  what  is  the  diameter  of  a  circle  whose  area  is  500  sq.  in., 
to  the  nearest  64th  of  an  inch?  Ans.  25^|  in. 

(4)  The  price  paid  for  a  certain  dye  was  $1.36  per  ounce.  This 
represented  a  dealer's  profit  of  33^%,  a  wholesaler's  profit  of 
\2\%,  an  importer's  profit  of  8f  %,  and  a  manufacturer's  profit 
of  20%;  what  was  the  cost  of  manufacturing,  after  allowing  4 
cents  for  handling,  packing,  etc.?  Give  result  to  the  nearest 
cent.  Ans.  66  cents  per  ounce. 

(5)  A  bill  for  rubber  hose  amounted  to  $235.40,  list  price;  dis- 
counts of  70,  30,  10  and  5%  were  allowed;  (a)  how  much  was 
actually  paid  to  settle  this  bill?  (6)  what  was  the  equivalent 
single  discount? 

/(a)  $42.27. 
AnS-    I  (6)  82.045%. 

(6)  A  sample  of  coal  shows  10.83%  ash;  if  the  weight  of  ashes 
obtained  in  one  day  is  1632  lb.,  actual  weight,  about  how  many 
long  tons  of  coal  were  burned? 

Ans.  6  T.  1629  lb.  =  6.727  T. 

(7)  A  firm  desires  to  pay  a  special  bonus  of  2%  on  the  com- 
missions earned  by  its  salesmen  when  the  sales  are  in  excess  of 
a  certain  fixed  amount;  but  the  2%  is  to  be  computed  on  the  sales 
(commissions)  after  deducting  the  special  bonus.  What  is  the 
actual  bonus,  expressed  as  a  per  cent?  Ans.  1.96+%. 

127 


128  ARITHMETIC  §1 

(8)  In  a  certain  mill  35  men  receive  42c'  per  hour,  64  receive 
7:'.r  per  hour,  15  men  receive  87|p  per  hour,  and  5  men  receive 
I1.12J  per  hour;  (a)  what  was  the  average  wage  per  hour  per 
man?  (b)  if  they  all  received  an  increase  of  \2\%,  what  was  the 
average  wage  per  hour  per  man? 

f  (a)  67.376  per  hr. 
I  (6)  75.79c  per  hr. 

(9)  By  making  certain  changes,  a  pulp  mill  increased  its  daily 
production  from  88  tons  to  95  tons.  The  total  cost  of  operation 
increased  25^,  and  the  price  was  increased  12|%;  what  was  (a) 
the  profit  per  cent  after  the  change,  and  (6)  what  was  the  gain 
or  loss  per  cent  in  profits,  if  the  profit  when  the  daily  production 
was  88  tons  was  18%?  4         /  (a)   14.65-  % 

I  (6)   18.6+  %  loss. 

(10)  How  many  gallons  are  equivalent  to  9.24  cu.  ft.? 

Ans.  69.12  gal. 

(11)  Add  5  yd.  2  ft.  7  in.,  3  yd.  1  ft.  8  in.,  4  yd.  9  in.  and  3 
yd.  2  ft.  10  in.  Ans.  3  rd.  1  yd.  4  in. 

(12)  The  sum  of  the  three  angles  of  any  plane  triangle  is  180°; 
if  two  of  the  angles  of  a  certain  triangle  are  36°  14'  43"  and  65° 
27'  13",  what  is  the  other  angle?  Ans.  78°  18'  4". 

(13)  Express  4.807  mi.  in  miles  and  lower  denominations. 

Ans.  4  mi.  6  fur.  18  rd.  1  yd.  11.52  in. 

(14)  Express  75°  18'  18"  as  a  decimal  part  of  360°. 

Ans.  0.2091805f. 

(15)  What  is  (a)  the  weight  in  ounces  of  57  c.c.  of  water? 
(6)  what  is  the  equivalent  volume  in  cubic  inches. 

Ans     {  {a)  2-0106+  °Z- 

I  (6)  3.4784+  cu.  in. 

(16)  Divide  26  mi.  6  fur.  22  rd.  3  yd.  2  ft,  6  in.  by  15. 

Ans.  1  mi.  6  fur.  12  rd.  2  ft,  11.6  in. 

(17)  Reduce  (a)  1  mi.  6  fur.  12  rd.  2  ft.  11.6  in.  to  inches; 
{b)  express  this  number  in  yards.  .         /  (a)  113,291.6  in. 

<  (6)  3146|*  yd. 


SECTION  2 

ELEMENTARY  APPLIED 
MATHEMATICS 

(PART  1) 


MATHEMATICAL  FORMULAS 


3t 


DEFINITIONS 

1.  A  mathematical  formula,  or,  more  simply,  a  formula,  is  an 
expression  composed  of  ordinary  arithmetical  numbers  and 
quantities  indicated  by  letters,  which  shows  at  a  glance  what 
operations  (addition,  subtrac- 
tion, multiplication,  division, 
powers,  and  roots)  are  required 
to  be  performed  in  order  to  ^-jy 
obtain  a  certain  desired  result. 

Roughly  speaking,  a  formula  is   U—i)— A 

a  short,  concise  expression  of  a  Fig.  l. 

rule,    law,    or    principle.      For 

instance,  refer  to  Fig.   1,  which  represents  a  cylinder  with  a 

round  hole  throughout  its  entire  length.     The  rule  for  finding 

the  weight  of  this  hollow  cylinder  may  be  stated  as  follows: 

Rale.— Multiply  the  sum  of  the  diameters  of  the  cylinder  and 
hole  by  their  difference;  multiply  this  product  by  the  length  of  the 
cylinder,  by  the  weight  in  pounds  of  a  cubic  inch  of  the  material  of 
which  the  cylinder  is  composed,  and  by  .7854,  all  measurements  to 
be  taken  in  inches.  The  final  product  will  be  the  weight  of  the 
cylinder  in  pounds. 

To  express  this  rule  by  a  formula, 

let  W  =  weight  of  cylinder  in  pounds; 

w  =  weight  in  pounds  of  a  cubic  inch  of  the  material  composing 

cylinder; 

1 


2  ELEMENTARY  APPLIED  MATHEMATICS  §2 

D  =  diameter  of  cylinder  in  inches; 

d    =  diameter  of  hole  in  inches; 

I     —  length  of  cylinder  in  inches; 
then  W  =  .7$rAwl{D  +  d)(D  -  d) 

This  last  expression  is  a  formula,  and  it  shows  at  a  glance  just 
what  operations  are  required  in  connection  with  the  quantities 
w,  I,  D,  d,  and  the  number  .7854  in  order  to  find  the  value  of  W, 
the  weight  of  the  cylinder.  All  that  is  necessary  in  order  to  use 
the  formula  is  to  substitute  in  it  the  values  of  the  quantities 
represented  by  the  letters.  Thus,  suppose  the  diameter  of  the 
cylinder  is  12  in.,  diameter  of  hole  is  8  in.,  length  of  cylinder  is  30 
in.,  and  that  it  is  composed  of  cast  iron,  a  cubic  inch  of  which 
weighs  .2604  pounds;  what  is  the  weight  of  the  cylinder?  Here 
D  =  12,  d  =  8,  I  =  30,  w  =  .2604;  substituting  these  values  for 
their  corresponding  letters  in  the  formula, 

W  =  .7854  X  .2604  X  30(12  +  8)  (12  -  8) 
or,  W  =  .7854  X  .2604  X  30  X  20  X  4  =  490.8+  lb. 

2.  When  two  or  more  letters  in  a  formula  are  written  with  no 
sign  of  addition  or  subtraction  between  them  or  when  a  letter 
follows  a  number,  multiplication  is  understood;  multiplication  is 
also  understood  when  a  letter  precedes  or  follows  a  sign  of  aggre- 
gation when  there  is  no  +  or  —  sign  between  or  when  two  dif- 
ferent signs  of  aggregation  follow  each  other;  therefore,  .7854u;Z 
(D  +  d)(D  —  d)  has  the  same  meaning  as  though  written  .7854 
X  w  X  I  X  (D  +  d)  X  (D  -  d).  Obviously,  the  sign  of  multi- 
plication cannot  be  omitted  between  two  numbers,  since  it  would 
not  then  be  possible  to  distinguish  between  the  numbers. 

3.  Positive  and  Negative  Quantities. — Before  going  further, 
it  is  necessary  to  make  a  distinction  between  quantities  that 
indicate  exactly  opposite  directions  or  meanings.  For  example, 
suppose  two  men  start  from  the  same  position,  one  walking  due 
north  and  the  other  due  south;  they  are  evidently  walking  in 
exactly  opposite  directions.  To  indicate  this  fact  mathematically, 
one  man  is  said  to  walk  in  a  positive  direction,  while  the  other  is 
said  to  walk  in  a  negative  direction.  Suppose  that  north  is  taken 
as  the  positive  direction;  then  if  one  man  walks  6  miles  due  north 
and  the  other  walks  6  miles  due  south,  the  first  man  is  said  to 
walk  -f  6  miles,  and  the  second  man  is  said  to  walk  —6  miles. 
Positive  quantities  are  always  indicated  by  the  plus  sign  and 
negative  quantities  by  the  minus  sign.     A  negative  quantity  is 


§2  MATHEMATICAL  FORMULAS  3 

just  as  real  as  a  positive  quantity;  it  simply  indicates  a  direction 
or  meaning  exactly  opposite  in  character  to  that  indicated  by  the 
positive  quantity. 

A  man's  income  may  be  considered  as  positive  and  his  expendi- 
tures as  negative.  If  going  up  hill  be  considered  positive,  going 
down  hill  will  be  negative.  When  the  mercury  in  a  thermometer 
is  above  zero,  the  reading  is  considered  positive;  when  it  is  below 
zero,  the  reading  is  negative;  hence,  65  degrees  above  zero  is 
written  +  65°,  and  12  degrees  below  zero  is  written  —  12°.  If  a 
push  be  considered  positive,  a  pull  will  be  negative.  Many 
other  instances  may  be  cited;  all  that  is  necessary  is  that  if  any 
particular  state,  meaning,  or  direction  be  considered  as  positive, 
the  state,  meaning,  or  direction  that  is  directly  opposite  in 
character  will  be  negative.  For  instance,  the  direction  in  which 
the  hands  of  a  clock  move,  called  clockwise,  may  be  considered 
as  positive;  then  if  the  hands  are  moved  backward,  or  counter- 
clockwise, this  direction  will  then  be  negative. 

4.  In  connection  with  arithmetical  problems,  it  is  always  pos- 
sible to  ascertain  which  of  two  numbers  is  the  larger;  or,  if 
several  numbers  are  under  consideration,  it  is  always  possible  to 
arrange  them  in  their  relative  order  of  magnitude.  But  when 
quantities  are  represented  by  letters,  it  is  seldom  possible  to 
arrange  them  in  this  manner,  and  it  frequently  happens  that  the 
subtraction  of  a  larger  quantity  from  a  smaller  may  be  indicated 
in  a  formula.  The  distinction  between  positive  and  negative 
quantities  becomes  of  great  importance  when  the  quantities  are 
represented  by  letters.  For  this  reason,  it  is  necessary  to  know 
how  to  add,  subtract,  multiply,  and  divide  positive  and  negative 
quantities  that  are  represented  by  letters. 

When  quantities  are  represented  by  letters,  they  are  called 
literal  quantities,  to  distinguish  them  from  those  represented  by 
figures  only  or  figures  combined  with  a  name  (concrete  numbers), 
which  are  called  numerical  quantities.  The  numerical  value  of 
any  quantity  is  its  value  when  expressed  by  figures.  Thus,  in 
Art.  1,  the  numerical  value  of  D  was  12;  of  d,  8;  of  w,  .2604,  etc. 

5.  Coefficients  and  Exponents. — It  was  stated  in  Art.  2  that 
an  expression  like  5a  means  5  X  a.  The  number  5  which  multi- 
plies a  is  called  the  coefficient  of  a;  the  coefficient  of  any  literal 
quantity  is  always  a  multiplier  of  the  quantity.  In  3^w,  ISapq, 
3(a  +  76),  etc.,  3^,  18,  3,  etc.  are  the  coefficients  of  m,  apq, 
(a  -f  76),  etc. 


4  ELEMENTARY  APPLIED  MATHEMATICS  §2 

It  is  frequently  convenient  to  represent  the  coefficients  by 
letters;  thus,  the  area  of  a  ciicle  is  expressed  by  the  formula 
A  =  irr-  in  which  w  represents  3.141592+  (usually  taken  as 
3.1416),  and  r  represents  the  radius  of  the  circle.  Jn  ISapq,  18a 
may  be  considered  as  the  coefficient  of  pq,  and  18ap  may  be 
considered  as  the  coefficient  of  q.  Here  18  is  the  numerical 
coefficient  of  apq,  while  a  and  ap  are  literal  coefficients  of  pq  and  q, 
respectively.  In  general,  however,  the  word  coefficient  refers 
only  to  the  numerical  coefficient.  If  no  numerical  coefficient  is 
given,  it  is  always  understood  to  be  1.  Thus,  the  coefficient 
of  mn  is  +1,  and  mn  =  -\-lmn;  the  coefficient  of  —  ay  is  —  1, 
and  —ay=  —  lay.  Note  that  when  no  sign  is  prefixed  to  a 
literal  expression,  it  is  always  understood  to  be  +;  thus,  18  apq 
=  +  ISapq.  The  minus  sign,  however,  is  never  omitted;  hence, 
the  numerical  coefficient  in  —236a;  is  —23. 

6.  Exponents  have  the  same  meaning  in  connection  with  literal 
quantities  that  they  have  when  used  with  numbers.  For  instance, 
an2  =  a  X  n  X  n;  -4a263  =  -4XaXaX6X6X6; 
8xy  =  8  X  x1  X  yl;  etc.  Note  particularly  that  any  quantity, 
whether  numerical  or  literal,  that  has  0  for  an  exponent  is  always 
equal  to  1;  thus,  m°  =  1,  56°  =  1,  etc.  The  reason  for  this  will 
be  explained  in  connection  with  division  of  literal  quantities. 

Exponents  may  also  be  negative,  in  which  case,  the  quantity 
affected  with  a  negative  exponent  is  always  equal  to  1  divided  by 
the  quantity  when  the  exponent  is  the  same,  but  positive;  thus, 

-i  1  r 

x~2  =  -v  —  15ay~3  = 3 — >  etc.     When  no  sign  is  prefixed  to 

x  y 

a  number  or  exponent,  it  is  always  understood  to  be  + ;  hence, 

2  =  +2,  x2  =  x+2,  etc.     1  divided  by  a  number  or  quantity  is 

called  the  reciprocal  of  that  number   or   quantity;   thus,    the 

reciprocal  of  2.5  is  jtvj  the  reciprocal  of  x2  is  ~w>  etc.     Therefore, 

any  quantity  affected  with  a  negative  exponent  is  equal  to 
the  reciprocal  of  that  quantity  affected  with  an  equal  positive 
exponent. 


OPERATIONS  WITH  LITERAL  QUANTITIES 

7.  Addition. — When  several  literal  expressions  are  connected 
with  one  another  by  the  signs  +  and  —  ,  their  sum  is  always 
understood.    For  instance,  in  the  expression  27a3  —  54a2#  + 


§2  MATHEMATICAL  FORMULAS  5 

36a?/2  -  Sy\  the  different  parts  connected  by  the  signs  +  and  - 
are  called  terms.  The  first  term,  27a3  is  understood  to  be  +27a3, 
the  second  term  is  -54a2y,  etc.  The  expression  is  equivalent  to 
+  27a3  +  (-54a2;/)  +  36a</2  4-  (Sy3). 

8.  Like  terms  are  those  having  the  same  literal  quantities 
affected  with  the  same  exponents,  regardless  of  the  coefficients; 
thus,  bay2  and  Say-  are  like  terms,  and  -bm  and  libra  are  also  like 
terms.  In  the  expression  of  Art.  7,  no  two  of  the  terms  are  alike, 
whence  they  are  called  unlike  terms.  It  might  be  thought  at 
first  that  the  second  and  third  terms  were  alike,  but  they  are  not, 
since  the  letters  do  not  have  the  same  exponents. 

9.  Like  terms  may  be  added,  but  unlike  terms  cannot  be  added 
—the  addition  of  unlike  terms  can  only  be  indicated  as  above. 

I.  When  adding  literal  quantities,  a  slightly  different  meaning 
is  given  to  the  term  addition  from  that  employed  in  arithmetic, 
because  all  numbers  used  in  arithmetic  are  positive.  Referring 
to  Fig.  2,  suppose  a  man  desires  to  walk  toward  the  point  B; 


4 


o 

+  3  +10  +15 


A 

Fig.  2. 


then  any  movement  that  takes  him  in  the  direction  to  the  right 
may  be  called  positive  or  4- ,  and  any  movement  that  takes  him 
in  the  opposite  direction  or  toward  the  left  will  be  negative  or  - , 
regardless  of  the  point  started  from.  Suppose  he  starts  from  A 
and  takes  6  steps  toward  B;  the  distance  advanced  is  then  4^6 
steps.  If  he  now  takes  9  steps  more  in  the  same  direction,  he  will 
advance  4-9  steps.  The  total  distance  advanced  toward  B  is 
evidently  4-15  steps;  therefore,  the  sum  of  4-6  and  4-9  is  +15. 
This  case  corresponds  to  ordinary  addition  in  arithmetic. 

II.  Suppose,  however,  instead  of  walking  toward  B;  he  had 
walked  toward  C,  starting  from  A,  as  before.  If  he  takes  first 
6  steps  and  then  9  steps  toward  C,  his  first  advance  is  -6  steps 
and  his  second  advance  is  -9  steps,  and  his  total  advance  is  -  15 
steps;  from  which  it  is  seen  that  -6  4-  (-9),  or  -6  -9,  =  -  15. 
That  is,  to  find  the  sum  of  two  negative  quantities,  add  them  and 
prefix  the  minus  sign. 

III.  Suppose  he  had  walked  first  6  steps  toward  B  and  then  9 
steps  toward  C;  he  would  evidently  stop  at  -3,  since  counting 


6  ELEMENTARY  APPLIED  MATHEMATICS  §2 

9  steps  to  the  left  from  +6  makes  him  pause  at  -3.  Therefore, 
+6-9  =  -3  or  -9  +  6  =  -3. 

IV.  Suppose  he  had  walked  6  steps  toward  C  and  then  9  steps 
toward  B;  he  would  evidently  stop  at  +3.  Therefore,  -6  +  9 
=  +3  or  +9  -  6  =  +3. 

From  the  foregoing,  it  is  seen  that  when  two  numbers  have  the 
same  sign,  their  suiyi  is  the  sum  of  the  two  numbers  prefixed  by  the 
common  sign;  but  when  the  numbers  have  unlike  signs,  the  sum  is 
equal  to  the  difference  of  the  two  numbers  -prefixed  by  the  sign  of  the 
greater  number. 

10.  To  add  two  like  terms  having  one  or  more  literal  quantities 
in  them,  all  that  is  necessary  is  to  add  their  coefficients  and 
prefix  to  the  sum  the  sign  of  the  greater.  For  instance, 
lax  +  Aax  =  llax;  —  Sax  +  lax  =  4ax;  12mn  —  19mn 
=  —  Imn;  etc. 

If  there  are  more  than  two  like  terms  to  be  added,  find  the  sum 
of  those  having  positive  and  those  having  negative  coefficients 
separately,  and  then  add  the  two  sums;  thus,  3a  —  8a  —  a 
+  6a  —  11a  +  4a  =  —la,  since  3a  +  6a  +  4a  =  13a,  —8a  —  a 
-  11a  =  -20a,  and  13a  -  20a  =  -la. 

To  show  that  the  sum  may  be  found  by  simply  adding  the 
coefficients,  consider  the  expression,  1c  +  4c  =  lie.  Suppose  c 
represents  5  inches;  then  1c  =  7  X  5  in.  =  35  in.,  4c  =  4  X  5  in. 
=  20  in.,  and  35  in.  +  20  in.  =  55  in.  But  1c  +  4c  =  lie  =  11 
X  5  in.  =  55  in.  In  other  words.  7  of  something  plus  4  of 
something  is  equal  to  7  +  4  or  11  of  something,  and  it  makes  no 
difference  what  this  something  is,  whether  it  is  1  in.  or  5  in.  or 
105  in. ;  all  that  is  necessary  is  that  the  terms  be  alike. 

11.  Subtraction. — Subtraction  may  be  defined  as  the  difference 
between  two  quantities;  it  may  also  be  defined  as  that  quantity 
which  when  added  to  the  subtrahend  will  give  the  minuend  for 
the  sum.  Thus,  the  difference  between  15  and  6  is  9,  and  9  is  the 
number  that  will  produce  15  for  the  sum  when  it  is  added  to  6; 
here  6  is  the  subtrahend  and  15  is  the  minuend.  The  second 
definition  is  the  better  one  to  use  in  connection  with  subtraction 
of  literal  quantities. 

Ilef erring  to  Fig.  2,  call  movement  toward  B  positive  or  +  and 
movement  toward  C  negative  or  — ,  as  before. 

I.  Suppose  two  men  to  start  from  A ;  let  one  man  walk  9  steps 
toward  B.  and  the  other  15  steps  toward  B;  how  far  are  they 


§2  MATHEMATICAL  FORMULAS  7 

apart?    The  distance  between  them  is  evidently    15  -  (+9) 
=  6,  since  9  +  6  =  15. 

II.  Suppose  they  had  walked  toward  C  the  same  number  of 
steps;  then  the  distance  between  them  is  —  15  —(  —  9)  =  —6, 
since  -9+  (-6)  =  -15. 

III.  Suppose  one  of  the  men  had  walked  9  steps  toward  B  and 
the  other  15  steps  toward  C;  then  the  distance  between  them 
is  —15  —  (+9)  =  —  24,  the  -24  meaning  that  the  man  at  +9 
would  have  to  take  —24  steps  to  place  him  at  —15.  Also  note 
that  9+  (-24)  =  -15. 

IV.  Suppose  that  one  of  the  men  had  walked  15  steps  toward  B 
and  the  other  9  steps  toward  C;  then  the  distance  between  them  is 
15  —  (-9)  =  24,  that  is,  the  man  at  -9  would  have  to  walk 
+24  steps  to  get  to  +15.     Also  note  that  -9  +  24  =  +15. 

It  will  be  observed  in  the  foregoing  four  cases  that  the  difference 
may  be  found  in  each  case  by  changing  the  sign  of  the  subtrahend 
and  proceeding  as  in  addition.  Thus,  in  I,  the  sign  of  the  subtra- 
hend is  +;  changing  it  to  —  and  adding,  15  —  9  =  +6.  In  II, 
-15  +  9  =  -6;  in  in,  -15  -9  =  -24;  and  in  IV,  15  +  9 
=  +  24. 

Therefore,  to  subtract  one  term  from  another,  change  the  sign  of 
the  subtrahend  and  proceed  as  in  addition.  For  example,  ISar 
-  (7ar)  =  liar;  12bd  -  (22bd)  =  -10bd;  6m2-  (-31m2) 
=  +37m2;  -226V  -(-346V)  =  +  126V 

12.  Multiplication. — The  general  rule  for  the  signs  in  multipli- 
cation is :  If  the  coefficients  of  two  terms  that  are  multiplied  have  like 
signs,  the  sign  of  the  coefficient  in  the  product  xcill  be  -\-;if  they  have 
unlike  signs  {one  +  and  the  other  — ) ,  the  sign  of  the  coefficient  in 
the  product  will  be  -.  Thus,  +3  X  +5  =  +15,  and  -3  X  -5 
=  +15;  also  -3  X  +5  =  -15,  and  +3  X  -5  =  -15. 

A  little  consideration  will  make  this  rule  clear.  Call  a  man's 
debts  negative  and  his  savings  positive;  then  any  tendency  to 
increase  the  debts  will  be  negative,  and  any  tendency  to  in- 
crease the  savings  or  to  make  the  debts  less  will  be  positive. 
Now  suppose  the  man  saves  3  dollars  each  week  for  5  weeks; 
at  the  end  of  5  weeks,  he  will  have  saved  3  X  5  =  15  dollars. 
The  3  dollars  and  the  15  dollars  are  necessarily  positive,  re- 
presenting savings;  the  factor  5  is  also  positive,  because  it  in- 
dicates that  3  dollars  was  saved  5  times.  Hence,  +  3  X  +  5 
=  +  15.     In  the  same  way,  +3X-5=-15,  because  this 


8  ELEMENTARY  APPLIED  MATHEMATICS  §2 

operation  indicates  that  3  dollars  was  spent  5  times,  the  total 
amount  spent  being  15  dollars.  Also,  —  3  X  +5  =  —  15, 
because  this  operation  indicates  that  a  debt  of  3  dollars  was  in- 
creased 5  times,  and  an  increase  must  be  considered  as  positive, 
the  total  increase  in  d-cbt  being  15  dollars.  Finally,  —  3  X  —  5 
=  -f-  15;  here  a  debt  of  3  dollars  is  decreased  5  times,  because, 
if  +  5  indicates  an  increase,  —  5  must  indicate  a  decrease;  hence, 
if  a  man's  debt  decreases,  the  value  of  his  assets  increase,  that 
is,  there  is  a  positive  change,  which  amounts  to  +  15  dollars. 
A  similar  line  of  reasoning  may  be  applied  to  any  similar  case 
involving  positive  and  negative,  or  opposite,  changes  or  quantities. 

The  product  must  contain  all  the  literal  quantities  in  both 
multiplicand  and  multiplier;  thus,  66  X  9ay  =  54a&?/,  the  coeffi- 
cient of  the  product  being  equal  to  the  product  of  the  coefficients 
of  the  factors.  Hence,  —  la-x  X  2ax3  =  —  lia2axxz  =  — 14 
a3x4.  In  other  words,  if  the  same  literal  quantity  occurs  in  both 
factors,  it  mill  hare  an  exponent  in  the  product  equal  to  the  sum  of 
the  exponetits  in  the  two  factors.  In  the  last  case,  the  exponent  of 
a  in  the  product  is  equal  to  2  -f-  1  =  3,  and  the  exponent  of  x  is 
equal  to  1  +  3  =  4.  This  is  always  the  case,  whatever  the 
exponents,  and  whether  positive  or  negative.  Thus,  c2z5 
X  c"4*-2  =  c~2z3,  since  2-4=  -2,  and  5-2  =  3;  also,  2.  Id-79 
X  3.4d2  =  7.14d2-79,  since  2.1  X  3.4  =  744,  and  2  +  .79  =  2.79. 

13.  Division. — The  general  rule  for  signs  in  division  is  very 
similar  to  that  for  multiplication.  If  the  dividend  and  divisor 
have  like  signs,  the  sign  of  the  quotient  is  +  ;  if  they  have  unlike 
signs,  the  sign  of  the  quotient  is  — .  To  find  the  quotient,  divide 
the  coefficient  of  the  dividend  by  the  coefficient  of  the  divisor, 
and  the  result  will  be  the  coefficient  of  the  quotient;  to  this  annex 
the  literal  quantities  in  both  dividend  and  divisor,  but  changing 
the  signs  of  the  exponents  in  the  divisor,  and  then  add  the  ex- 
ponents of  like  quantities.  If  any  exponent  then  becomes  0,  the 
value  of  the  quantity  having  that  exponent  is  1,  and  it  thus  dis- 
appears from  the  quotient.  For  example,  48a263c  ■*■  —  12a26 
=  -4b2c,  since  48  ■*■  -12  =  -4,  and  a2b3ca-2b~l  =  a°b2c  =  162c 
=  62c. 

The  proof  for  the  law  of  signs  in  division  is  simple:  As  in 
arithmetic,  the  product  of  the  divisor  and  quotient  is  the  divi- 
dend (no  remainder  being  considered).  Letting  d  represent  the 
dividend,  p,  the  divisior,  and  q  the  quotient,  d  =  p  X  q.  If  d 
and  p  are  both  positive,  q  will  also  be  positive,  because  +  p 


§2  MATHEMATICAL  FORMULAS  9 

X  +  q  =  +  pq  =  +  d;  also,  if  d  and  p  arc  both  negative,  q 
will  be  positive,  because  —  p  X  -\-  q  =  —  pq  =  —  d.  Further, 
if  d  is  positive  and  p  negative,  q  is  negative,  because  —pX—q 
=  +  pq  =  +  d;  also  if  d  is  negative  and  p  positive,  q  is  negative, 
because  -\-  p  X  —  q  =  —  pq  =  —  d.  Therefore,  if  the  signs  of 
the  dividend  and  divisor  are  alike,  the  quotient  is  positive;  if 
they  are  unlike,  the  quotient  is  negative. 

In  practice,  the  exponents  of  the  literal  quantities  in  the  divisor 
are  subtracted  from  the  exponents  of  like  quantities  in  the  divi- 
dend; this  produces  the  same  result  as  changing  the  signs  of  the 
exponents.  Thus,  in  the  last  example,  the  exponent  of  a  in  the 
quotient  is  2  —  2  =  0;  of  b,  3  —  1  =  2;  and  as  c  does  not  occur 
in  the  divisor,  it  goes  into  the  quotient  unchanged. 

To  prove  that  any  number  having  0  for  an  exponent  is  equal  to 
1,  let  n  represent  any  number  or  quantity,  then  n  -S-  n  =  1;  but 
n  -4-  n  =  nl~l  =  n°.  Since  the  only  way  that  0  can  be  obtained 
for  an  exponent  is  to  divide  a  number  or  quantity  by  itself,  it 
follows  that  any  number  or  quantity  having  0  for  an  exponent  is 
equal  to  1.     Similarly,  n5  -5-  n5  =  n5-5  =  n°  =  1. 

14.  Operations  with  Polynomials. — An  expression  consisting 
of  but  one  term  is  called  a  monomial  (the  prefix  mo  is  a  contrac- 
tion of  mon  meaning  one);  an  expression  consisting  of  but  two 
terms  is  called  a  binomial  (bi  means  two);  an  expression 
consisting  of  more  than  two  terms  is  called  a  polynomial  (poly 
means  many). 

Before  performing  any  of  the  operations  of  addition,  subtraction, 
multiplication,  or  division,  it  is  always  advisable  to  arrange  the 
polynomials  according  to  the  descending  powers  of  one  of  the 
letters;  this  is  done  by  writing  first  that  term  containing  the 
highest  power  of  the  letter  chosen,  then  the  term  containing  the 
next  highest  power,  etc.  until  all  the  terms  have  been  written. 
For  example,  arrange  2x3y  -  2axif  +  xb  -  ?/4  +  x2y2  -  a2x* 
according  to  descending  powers  of  x.  The  term  containing  the 
highest  power  of  x  is  xb,  that  containing  the  next  highest  power  of 
x  is  —a2x4,  etc.  consequently,  the  polynomial  arranged  according 
to  descending  powers  of  x  is  x5  -  a2x*  +  2xhj  +  x2y2  -  2axy3 
—  y*.  If  it  were  desired  to  arrange  the  polynomial  according  to 
descending  powers  of  y,  the  result  would  then  be  -y*  —  2axy3 
_|_  xty2  _|_  2xhj  —  a2xA  +  x5.  Note  that  in  the  first  polynomial 
as  arranged,  the  last  term  does  not  contain  x,  and  in  the  second 
arrangement,  the  last  two  terms  do  not  include  y. 


10  ELEMENTARY  APPLIED  MATHEMATICS  §2 

15.  To  add  two  or  more  polynomials,  first  arrange  the  one 
containing  the  most  terms  acording  to  the  descending  powers  of 
one  of  the  letters;  then  place  under  this  the  other  polynomials, 
with  like  terms  in  the  same  columns,  and  add  each  column 
separately  as  in  addition  of  monomials. 

Example  1.— Find  the  sum  of  x2  -  Sxy  +  if-  +  x  +  y  -  1,  2x2  +  4xy 

—  3y2  -  2x  -  2y  +  3,  3x2  -  5xy  -  4y2  +  3x  +  4y  -  2,  and  6x2  +  lOxy 
+  5y2  +  x  +  y. 

Solution. — Arrange  the  first  polynomial  according  to  descending  powers 
of  x,  and  then  place  under  it  the  other  polynomials  with  like  terms  in  the 
same  columns.  When  adding  polynomials,  it  is  customary  to  begin  at  the 
left,  instead  of  at  the  right,  as  in  arithmetic,  and  this  can  be  done  because 
there  is  never  anything  to  carry  from  one  column  to  the  next. 
The  sum  of  the  coefficients  in  the  first  column  is  12,  and  the  first  term  of 
the  sum  is  12x2. 

x2  -    Sxy  +    x  +    y2  +  y    -  1 
2x2  +    4xy  -  2x  -  Zy2  -  2y  +  3 
3x2  -    5xy  +  3x  -  4y2  +  4y  -  2 
6x2  +  lOxy  +    x  +  5y2  +    y 
12x2  +    6xy  +  3x  -    y2  +  4y.     Ans. 

The   sum   of   the   coefficients   in   the   second    column   is    10  +  4—5 

—  3  =  6,  and  the  second  term  of  the  sum  is  6xy.  The  sum  of  the  coeffi- 
cients in  the  third  column  is  1+3  +  1— 2=3,  and  the  third  term  of  the 
sum   is  3x.     The  sum  of  the  coefficients  in  the  fourth  column  is  5  +  1 

—  4  —  3  =  —  1,  and  the  fourth  term  of  the  sum  is  —  y2.  The  sum  of  the 
coefficients  in  the  fifth  column  is  1+4  +  1— 2  =  4,  and  the  fifth  term 
of  the  sum  is  4y.     The  sum  of  the  numbers  in  the  sixth  column  is  3  —  2 

—  1=0,  and  the  entire  sum  is  12x2  +  6x,y  +  3x  —  y2  +  4y. 

Example  2. — Find  the  sum  of  4a  —  5b  +  3c  —  2d,  a  +  b  —  4c  +  5d,  3a 

—  lb  +  6c  +  4d,  and  a  +  4b  -  c  -  7d. 

4a  —  5b  +  3c  —  2d  Solution. — Since  the  letters  in  all  these  polynom- 

a  +    b  —  4c  +  5d  ials  have  the  same  exponents,  1,  arrange  them  accord- 

3a  —  76  +  6c  +  4d  ing  to  the  order  of  their  letters,  i.e.,  alphabetically. 

a  +  4b  —    c  —  Id  The  sum  is  then  found  in  the  same  manner  as  in 

9a  _  75  +  4c.     Ans.  Example  1. 

16.  To  subtract  one  polynomial  from  another,  arrange  the 
minuend  according  to  descending  powers  of  one  of  the  letters, 
place  the  like  terms  of  the  subtrahend  under  it  in  the  same  columns, 
and  subtract  each  term  of  the  subtrahend  from  the  term  above  it 
in  the  minuend,  as  in  subtraction  of  monomials.  If  the  subtra- 
hend contains  a  term  not  in  the  minuend,  change  its  sign  and 
write  it  in  the  remainder;  and  if  the  minuend  contains  a  term  not 
in  the  subtrahend,  write  it  in  the  remainder  with  its  sign 
unchanged. 


§2  MATHEMATICAL  FORMULAS  11 

Example  1.— From    4x3  -  2x2  +  3x'  -  1  +  7x    subtract    6x  +  1  -  x* 
+  2x3  -  2x2. 
Solution. — Arranging  the  minuend  according  to  descending  powers  of  x, 

and  writing  the  subtrahend  under  it  with 

like  terms  in  the  same  columns,  begin  at 

3x4  +  4x3  -  2x2  +  7x  -  1  the  left  and  subtract  each  term  of  the 

-x4  +  2x3  -  2x2  +  6x  +  1  subtrahend  from  that  above  it  in  the 

4x4  +  2x3  +    x  -  2.     ylns.minuend.     Since    3  -  ( -  1)  =  4,     the 

first  term  in  the  remainder  (difference) 
is  4x4.     The  second  term  of  the  differ- 
ence is  evidently  2x3,  the  third  term  is  0,  the  fourth  term  is  x,  and  the  coef- 
ficient of  the  fifth  term  is    —  1  —  (-fl)  =  —  1  —  1  =  —  2.     The  final 
result  obtained  for  the  difference  is  4x4  +  2x3  -f-  x  —  2. 

Example    2.— From     8ow5  -  13a3m3  +  18a4m2  -  29     subtract     8am5 
+  4a2m4  +  14a4m2  -  24a5m  -  38. 

Solution. — The  arrangement  according  to  descending  powers  of  m,  and 
allowing  for  missing  powers  in  both  subtrahend  and  minuend  is  shown  below. 
8ams  -  13a3m3  +  18a4m2  -  29 

8am5  +  4a2m4 +  14a4m2  -  24a5m  -  38 

-  4a2r«4  -  13a3m3  +  4a4m2  +  24a5m  +  9.  Ans. 
As  there  is  no  term  in  the  minuend  over  the  second  term  of  the  subtrahend, 
change  its  sign  and  write  it  in  the  difference.  There  is  no  term  in  the 
subtrahend  under  the  second  term  of  the  minuend;  hence,  bring  it  down  into 
the  difference  as  it  stands.  There  is  no  term  in  the  minuend  over  the 
fourth  term  of  the  subtrahend;  hence,  change  its  sign  and  bring  it  down  into 
the  difference.  The  rest  of  the  work  is  evident,  and  the  difference  sought 
is  -4a2m4  -  13a3m3  +  4a4m2+24a6m  +  9.  That  the  result  as  obtained 
is  correct  may  be  proved  by  adding  the  difference  to  the  subtrahend,  using 
the  work  as  it  stands;  the  sum  is  the  minuend. 

17.  To  multiply  a  polynomial  by  a  monomial,  simply  multiply 
separately  each  term  of  the  polynomial  by  the  multiplier.  It  is 
customary  to  arrange  the  multiplicand  according  to  the  descend- 
ing powers  of  one  of  the  letters  and  begin  the  multiplication  with 
the  term  containing  the  highest  power,  i.e.,  begin  at  the  left. 

Example. — Multiply  5x3  -  2ax2  +  7a2x  -  14a3  by  4a2x3. 

Solution. — The   multiplicand   is   already   arranged   according   to   the 

descending  powers  of  x.     Multiplying  each  term  separately  by  4a2x3,  the 

5x3  -  2ax2  +  7a2x  -  14a3  product  is  20a2x>  -  8a3x5  +  28 

4a2j.3  a*x*  -  56a5x3.     The  product  is 

20a2x6-8a3x5  +  28a4x4-56a5x3.     Ans.     ^sily  found  in  this  case  without 

placing  the  multiplier  under  the 
multiplicand,  but  this  has  been  done  in  the  margin  to  show  how  the  work  is 
arranged.  Beginning  at  the  left  and  taking  each  term  of  the  multiplicand 
in  succession,  5x3  X  4a2x3  =  20a2x6,  -  2ax2  X  4a2x3  =  -  8a3x5,  etc. 

18.  To  multiply  a  polynomial  by  a  binomial  or  another  polyno- 
mial,  arrange  both   multiplicand   and  multiplier  according  to 


a    -    3b 

4a»  -    5aJ6  +    6a62 

-  12a26  +  15a6*  -  186» 


12  ELEMENTARY    \PPLIED  MATHEMATICS  §2 

descending  powers  of  one  of  the  Letters,  placing  the  first  term  of 

the  multiplier  under  the  first  term  of  the  multiplicand.  Then 
multiply  each  term  of  the  multiplicand  by  the  first  term  of  the 
multiplier;  the  result  will  be  the  first  partial  product.  Multiply 
each  term  of  the  multiplicand  by  the  second  term  of  the  multiplier; 
the  result  will  be  the  second  partial  product,  which  is  written 
under  the  first  partial  product  with  like  terms  under  each  other. 
Proceed  in  this  manner  until  all  the  terms  of  the  multiplier  have 
been  used.  Now  add  the  partial  products,  and  the  sum  will  be 
the  entire  product. 

Example  1.— Multiply  4o*  -  5ab  +  662  by  a  -  36. 

SOLUTION. — Multiplying    the    multiplicand    by    a,    the    product   is    4a» 

—  5a26  +  6a62,  which  is  written  under 
the  multiplier.     Now  multiplying  by 

4ai  _    5a£    _j_  e^j  -  36,  the  product'of  4a2  and  —  36  is 

—  12a26,  which  is  written  in  the  same 
column  as  the  —  5a26  of  the  first  par- 
tial product.  Then  -  5ab  X  -  36 
=  15a62,  which  is  written  in  the  col- 

4a'  -  17a*6  +  21a6*  -  186s.     Ana.      umn  containing    6a62.     66s  X  -  36 

=   —  1S6S,  which  is  written  to  the 
right  of  the  preceding  product.    Add- 
ing the  two  partial  products,  the  entire  product  is  4«3  —  17a26  +  21a6!  —  1863. 

Example  2.— Multiply  D  +d  by  D-d. 

Solution. — The  work  is  shown  in  the  margin,  and  should  be   evident 
without  any  special  explanation.     The  entire  pro- 
duct is  D2  —  d2.     It  is  to  be  noted  that  D  and 
D  -\-  d  d  are  to  be  treated  as  though  they  were  two  different 

D  —  d  letters.     This  result  shows  that  the  product  of  the 

D1  +  Dd  sum  and  difference  of  two  numbers  or  quantities  is 

—  Dd  —  d1  equal  to  the  difference  of  their  squares,  and  the  for- 

j)i  —  d*.     ,4ns.     mula  in  Art.  1  may  be  written  .4  =  .7S54ui  {D-  —  d2). 

In    the    example    that    is    given    in     connection 
with    this  formula,    D  =  12  and  d  =  8;  D*  =  122 
=  144  and  d1  =  S1  =  64;  144  -  64  =  SO.     But  D 
+  d  =  12  -t-  8  =  20;    D  -  d  =  12  -  S  =  4;    (fl  +  d)    (D  -  d)  =  20  X  4 
=  80,  the  same  result  in  either  case. 

l'.\  wu'i.r.  3.— Multiply  z*+   2x3  +  j2  -  Ax  -  11  by  5j2  -  2x  +  3. 
Solution. — The  work  is  shown  below  and  requires  no  special  explanation. 
x*  +    2x3  +    z*  -    4.r  -  11 

&r»  -    2x   +  3 

5x«  +  10js  +  5x*  -  20xJ  -  55x* 

-    2j5  -  4x<  -    2j3  +    Sx*  +  22* 

3x<  +    fo1  +    3x*  -  \2x  -  33 

5x*  +    8x>  +  4x4  -  16<r*  -  44x2  +  lOx  -  33.     .Ins. 


§2  MATHEMATICAL  FORMULAS  13 

It  will  be  observed  here   that  when  the  multiplicand  and  multiplier  are 

arranged  according  to  the  descending  powers  of  some  letter,  each  partial 

product  begins  one  place  farther  to  the  right  than  the  preceding  partial 

product,  and  that  all  terms  of  the  partial  products  follow  in  regular  order. 

Example  4. — Suppose  the  multiplier  in  the  last  example  had  been  x2 

—  2x  +  3;  what  would  the  product  have  been? 

Solution. — The  work  is  shown  herewith.     This  example  has  been  selected 

x4  +  2x3  +    x2  -  4x    -  11 

x2  -  2x    +  3 

x6  +  2x6  +    x4  -  4x3  -  llx2 

-  2x5  -  4x4  -  2x3  +    8x2  +  22x 

3x4  +  6x3  +    3x2  -  12x  -  33 

x13  +  lOx  -  33.     Ans. 

to  show  how  terms  will  sometimes  disappear  in  multiplication,  the  product 
in  this  case  being  x6  +  lOx  —  33,  the  terms  containing  x5,  x4,  x3,  and  x2 
having  disappeared. 

19.  To  divide  a  polynomial  by  a  monomial,  arrange  the  polyno- 
mial according  to  the  descending  powers  of  one  of  the  letters  and 
divide  each  term  separately  by  the  divisor.  Thus,  to  divide 
21a462  -  12a363  +  24a264  -  33a65  by  3ab2,  arrange  the  work  in 
exactly  the  same  manner  as  for  short  division  in  arithmetic. 

3ob2)21a'b:!  -  12a3b3  +  24a2b4  -  330^ 

7o3   -    4a2b    +    Sab2    -lib3.    Ans. 
Beginning  with  the  first  term  of  the  dividend  and  dividing  it  by 
3o62,  the  quotient  is  7a3,  which  is  the  first  term  of  the  quotient 
sought.     The  other  terms  are  found  in  the  same  way,  the  quotient 
being  7a3  -  4a2b  +  Sab2  -  1163. 

20.  To  divide  a  polynomial  by  a  binomial  or  by  a  polynomial, 
arrange  the  dividend  and  divisor  according  to  descending  powers 
of  one  of  the  letters,  placing  them  in  the  same  relative  positions 
as  for  long  division  in  arithmetic.  Divide  the  first  term  of  the 
dividend  by  the  first  term  of  the  divisor,  and  the  result  will  be  the 
first  term  of  the  quotient;  multiply  the  divisor  by  the  first  term 
of  the  quotient  and  subtract  the  product  from  the  dividend. 
Divide  the  first  term  of  the  remainder  just  found  by  the  first 
term  of  the  divisor,  and  the  result  will  be  the  second  term  of  the 
quotient,  which  multiply  into  the  divisor  and  subtract  the  prod- 
uct from  the  remainder  fiist  found.  Proceed  in  this  manner  until 
a  remainder  of  0  is  obtained  or  until  the  first  term  of  the  last  re- 
mainder is  of  lower  degree  than  the  first  term  of  the  divisor,  in 
which  case,  write  the  remainder  as  the  numerator  of  a  fraction 
whose  denominator  is  the  divisor. 


-  aft2  +  b3     (B) 

-  ab2  +  b3 


14  ELEMENTARY  APPLIED  MATHEMATICS  §2 

Example  1. — Divide  a3  —  2ab2  +  b3  by  a  —  b. 

Solution. — Arranging  the  dividend  and  divisor  according  to  the  descend- 
ing  powers  of  a,  place  the  divisor  to  the  right  of  the  dividend,  with  a 

curved  line  between,  and  draw  a  line 
a3  _  2ab-  +  b3  (a  —  b  under  the  divisor  to  separate  it  from 

a3  -  a2b  ai  +  ab  _  &j.     Am.     the    <iuotient-     Then>    «3  *  a  =  a'> 

2b  _  ,  m   i   i)     /  j\  *be    first    term    of    the    quotient. 

ml  _        ,  j  Multiplying   the   divisor  by  a2  and 

subtracting  the  product  from  the  div- 
idend, the  remainder  is  shown  at  {A), 
which  is  arranged  according  to  de- 
0        0       (C)  scending    powers   of   a.     Then,    a"b 

-j-  a  =  ab,  the  second  term  of  the 
quotient.  Multiplying  the  divisor  by  ab  and  subtracting  the  product  from  the 
remainder  (A),  the  new  remainder  isshown  at  (B).  Then,  —ab2  -5-  a  =  —6*, 
the  third  term  of  the  quotient.  Multiplying  the  divisor  by  —  b2  and  sub- 
tracting the  product  from  the  remainder  at  (B),  the  difference  is  0,  showing 
that  the  division  is  exact.     The  quotient  sought  is  a2  -f-  ab  —  b2. 

Example  2.— Divide  64x5  -  486a5  by  2x  -  3a. 

Solution. — The  work  is  shown  below  and  requires  no  special  explanation. 
64x5  -  486a5  (2x  -  3a 

64z5  -    96ax4  32x<  +  48ax3  +  72a2x2  +  108a3x  +  162a4.     Ans. 
96ax<  -  486a5 
96ax<  -  144a2x3 


144a2x3  -  486a5 
144a2x3  -  216a3x2 


216a3x2  -  486a6 

216a3x2  -  324a«x 

324a4x  -  486a5 
324a<x  -  486as 

That  the  quotient  as  obtained  is  correct  may  be  proved  by  multi- 
plying it  by  the  divisor;  the  product  will  be  the  dividend. 

Example  3.— Divide  9n4  +  Hen2  +  17  c2n  by  3n2  +  4c. 
Solution. — To  obtain  the  second  term  of  the  quotient,  note  that 
9n*  +  lien2  +  I7c2n   (3n*  +  4c 

°*±1**  3n^  +  c^+i>.    An, 

-  en2  +  \7r";t  9n2  +  12c 

-  en2  -  jc2 


17  c*n  +  jc2        51c«w  +  4  c2        c'(51n  +  4) 
3n2  +4c  On1  +  12c      =    9n2  +  12c 

—  en2  -j-  3n2  =  —  \c.  Note  further  that  the  exponent  of  n  in  the  first 
term  of  the  second  remainder,  17c2/;,  is  smaller  than  the  exponent  of  n  in 
the  divisor;  hence,  the  remainder  is  said  to  be  of  lower  degree  than  the 
divisor,  and  the  division  ceases,  the  remainder  being  written  as  the  numerator 
of  a  fraction  whose  denominator  is  the  divisor.  The  numerator  contains  the 
fraction  £ ;  to  get  rid  of  it,  multiply  both  numerator  and  denominator  of  the 


§2  MATHEMATICAL  FORMULAS  15 

entire  fraction  by  3  (as  shown  in  arithmetic,  this  does  not  alter  the  value  of 

51c2?i  +  4c2 
the  fraction),  and  the  result  is  ~Qn2  +  y2c  '     lt  is  readily  seen  that  51chi 

_(_  4C2  _,  c2(5in  _(_  4)^  since  if  the  parenthesis  be  removed  and  the  terms 
within  it  are  multiplied  by  c2,  the  product  will  be  the  binomial  51c2n  +  4c2. 

Therefore,  the  quotient  is  3rt2  -  \c  +  C*[5?n,  *t^ . 

9n2  +  12c 

21.  Signs  of  Aggregation. — The  signs  of  aggregation  are  used 
much  more  freely  in  connection  with  literal  quantities  than  with 
numerical  quantities,  the  reason  being  that  numerical  quantities 
can  be  readily  combined  to  form  a  single  number,  which  is  not 
possible  with  literal  quantities,  unless  the  terms  are  alike.  The 
rule  given  in  arithmetic  for  removing  the  signs  of  aggregation 
applies  to  both  numerical  and  literal  quantities.  Thus,  a  + 
[b  +  (c-  d)]  =  a  +  [b  +  c  -  d]  =  a  +  b  +  c  -  d.  Here,  as  in 
arithmetic,  when  one  sign  of  aggregation  includes  another,  the 
inner  one  is  removed  first.  This  is  particularly  advisable  when 
some  of  the  signs  are  negative.  Thus,  a  —  [b  —  (c  —  d)] 
=  a  —  [b  —  c  -\-  d]  =  a  —  b  +  c  —  d. 

Example   1. — Remove  the  signs  of  aggregation   from  2x  —  \3y  —  [Ax 

-  (5y  -  6x)]}. 

Solution.— 2x  -  {3y  -  [4x  -  (5y  -  Gx))}  =  2x  -  [Zy  -  [Ax  -  by 
+  6x}\   =  2x  -  {3y  -  [10z  -  5^]}   =  2x  -  {Zy  -  10x  +  5y\   =  2x  -  {8y 

—  10x}  =  2x  —  8y  +  lOx  =  \2x  -  8y.  First  remove  the  parenthesis;  since 
the  quantities  enclosed  by  it  are  subtracted  from  4x,  their  signs  must  be 
changed,  thus  making  the  expression  within  the  brackets  Ax  —  5y  +  6x, 
which  is  equal  to  lOx  -  5y.  Next  remove  the  brackets,  obtaining  for  the 
expression  within  the  brace  3y  —  lOx  +  5y,  which  is  equal  to  Sy  —  lOx. 
Now  removing  the  brace,  the  original  expression  has  reduced  to  2x  —  8y 
+  lOx,  which  is  equal  to  12x  —  8y.     Ans. 

Example  2. — Remove  the  signs  of  aggregation  from 
5m{(a  -  b)[a2  -  Ab(a*  -  b2))\ 

Solution.—  5m{(a  -  fe)[a2  -  Ab(a2  -  b2)]}  =  5m{  (a  -  b)[a2  -  Aa*b 
+  Ab3]}  =  5m{a3  -  Aa3b  +  Aab3  -  a2b  +  Aa2b2  -  4fc4}  =  bma3  -  20ma3b 
—5ma2b  +  20ma2b2  +  20mab3  —  20rnb*.  Ans.  Having  removed  the  paren- 
thesis, the  polynomial  within  the  brackets  is  multiplied  by  the  binomial, 
a  —  b,  and  the  brackets  are  removed.  Now  removing  the  brace  and 
multiplying  the  last  product  by  5m,  clears  the  expression  of  all  the  signs  of 
aggregation. 

22.  When  two  or  more  terms  of  an  expression  have  a  common 
factor,  the  terms  may  be  included  in  a  parenthesis  or  other  sign 
of  aggregation  and  the  common  factor  placed  outside  as  a  multi- 
plier. For  instance,  the  polynomial  x2  +  2ax  -f  a2  may  be 
written  x(x  +  2a)  -f-  a2  or  x2  4-  a  (2x  +  a),  if  desired.     When  the 


16  ELEMENTARY  APPLIED  MATHEMATICS  §2 

first  term  within  the  parenthesis  has  the  minus  sign  before  it,  as  in 
x2  —  2ax  +  a2  =  x2  +  a(  —  2x  +  a),  it  is  desirable  to  have  the 
first  term  positive,  and  this  may  be  done  in  the  present  case  by- 
changing  the  order  of  terms,  writing  the  expression  x2  +  a{a 

-  2x).  If,  however,  it  is  not  desired  to  change  the  order  of  terms 
or  if  both  terms  are  negative,  -place  the  minus  sign  outside  the 
parenthesis  and  change  the  signs  of  all  the  terms  ivithin  it;  in  such 
case,  the  last  expression  becomes  x2  —  a(2x  —  a).  That  this  is 
correct  will  be  clear  when  it  is  noted  that  the  expression  as  it 
stands  means  that  a(2x  —  a)  is  to  be  subtracted  from  x2,  and  in 
subtraction,  the  signs  of  the  subtrahend  are  changed,  thus  making 
the  expression  x2  —  2ax  +  a2,  the  original  form  when  the  paren- 
thesis is  removed. 

Suppose  it  were  desired  to  divide  x3  —  x2y  +  xy2  —  yz  by 
x  —  y.  This  may  be  done  in  the  regular  way,  but  a  somewhat 
easier  method  in  this  case  is  the  following:  x3  —  x2y  +  xy2  —  y3 
=  x2(x  —  y)  +  y\x  —  y)  =  (x  —  y)(x2  -f  y2),  and  this  last  ex- 
pression divided  by  x  —  y  gives  x2  -f-  y2  for  the  quotient.  That 
x2(x  —  y)  +  y2(x  —  y)  =  {x  —  y)  (x2  -\-y2)  is  evident,  because  x  —  y 
is  a  factor  common  to  both  terms,  and  if  the  parenthesis  be 
removed  from  (x2  +  y2),  the  expression  reduces  to  the  preceding 
one. 

If  more  than  one  of  the  signs  of  aggregation  are  used,  the  inner 
one  is  used  first  and  the  terms  enclosed  by  it  are  treated  as  a 
single  term  when  employing  the  next  sign.     For  example,  3m3 

-  \2m2n  -  6mn2  -  18w3  =  3m3  -  12?n2n  -  6n2(m  +  3n)  =3ra3 

-  6»[2ma  +  n(m  +  3n)]  =  3{m3  -  2n[2m2  +  n(m  +  3n)]|. 
Note  that  the  signs  of  the  terms  within  the  parenthesis  are  not 
changed  when  the  brackets  are  written,  the  entire  expression, 
n(m  +  3n),  being  treated  as  though  it  were  a  single  letter. 

23.  The  Signs  of  a  Fraction^ — A  fraction  has  three  signs,  one 
for  the  numerator,  one  for  the  denominator,  and  one  for  the 
entire  fraction,  the  latter  showing  whether  the  fraction  is  to  be 
added  or  subtracted.  Thus,  let  a  be  the  numerator  and  b  the 
denominator,  then  the  fraction  may  have  one  of  the  following 

,  +a       a  —a  a     .    +a  a         +a  a 

forms:  ++6  =  fc,    +  ^  =   -g.  +  -_r6  =   -y  -  ^  -  -y 

—  a     a         +a    _a         —a  _    _a 
""  +6  =  Z>  ~  Z-fr  ~ V  ""  ^-b  ==  ~ b' 

To  prove  that  these  results  are  correct,  apply  the  law  that  if 
an>-  quantity  be  multiplied  and  divided  by  the  same  number  or 


§2  MATHEMATICAL  FORMULAS  17 

quantity  its  value  is  unchanged;  thus,  3  X  -1  =  -3,  and  -3 
-J-  -1  =  3.  Now  remembering  that  to  divide  a  fraction,  its 
numerator  may  be  divided  or  its  denominator  multiplied  by  the 

divisor,   -1  X  (+^)     -    -l    =     -=±frl  -_•._, 

,    —  a  -. —  1  a     a 

—  i  i  jf,     ~  —    •"&  ~  5*    e*c-     Now    since    multiplying   by 

-  1  changes  the  sign  of  the  fraction  and  dividing  by  -  1  changes 
the  sign  of  the  numerator  or  denominator,  as  the  case  may  be,  it 
follows  that  if  two  of  the  three  signs  of  the  fraction  are  changed,  the 
value  of  the  fraction  is  not  altered;  but,  if  one  sign  only  or  all  three  are 
changed,  the  value  is  altered. 

3^  2la 

Suppose  the  fraction  has  the  form = = — ;  to  make  the 

x-  —  a2      ' 

sign  before  the  first  term  of  the  numerator  +  ,  change  the  sign 
of  each  term  of  the  numerator  and  the  sign  of  the  fraction, 
,  .  .  .           3z  +  11a      mi 
obtaining _       .      I  he  same  result  might  have  been  ob- 

tained  by  changing  the  signs  of  both  numerator  and  denominator, 

,  .   .  .      3x  +  11a   ,         .  .     , 

obtaining       2_x2  1  but  tms  changes  the  order  of  the  terms  in  the 

denominator,  which  is  not  usually  desirable. 

—-4-  24r2 
Similarly,      2  "  C  +  i8°*  '  48°' 


4n2  +  In  -  c~8n2  +  14n  -  2c~  ~8n2  +  14n  -  2c 
Here,  to  get  rid  of  the  fraction  in  the  numerator,  both  numer- 
ator and  denominator  are  multiplied  by  2  (this  does  not  alter 
the  value  of  the  fraction) ;  then  the  sign  before  the  fraction  and 
the  signs  of  the  terms  in  the  numerator  are  changed.  If  the 
signs  of  the  numerator  and  denominator  are  changed,  instead  of 
the  sign  of  the  numerator  and  the  sign  of  the  fraction,  the  result 

•11  }  >  c-48c2  _  c-48c2 

W1     M  -8n2  -  14n  +  2c  ~  2c-14n-8n2- 


EXAMPLES 

(1)  Find  the  sum  of  a2  -  b3  +  3a*b  -  Bab2,  3a2  +  3/>3  -  3nb2  -  4a% 
a3  +  b3  +  3a2b,  2a3  -  4b3  -  5a&s,  (!«=/>  -  3a3  +  10a.>2,  and  2b3  -  7a2b 
+  4ab2.  Ans.  4a2  +  a2b  +  ab2  +b3. 

(2)  From  3w3  -  2m2  —  m  —  7  subtract  2m3  -  3/n2  +  m  +  1. 

Ans.  ?n3  +  m2  —  2m  —  8. 
2 


18  ELEMENTARY  APPLIED  MATHEMATICS  §2 

(3)  Add  a2  -  Say  +  y2  +  a  +  y  -  1,  iay  -  2a  +  2a2  -  3y2  -  2y  +  3, 
3a2  -  bay  -  Ay2  +  iy  -  2,    and    5?/2  +  10ay  +  6o2  +  y  +  a. 

Ans.  12a2  +  6ay  —  y2  +  \y. 

(4)  Subtract     a  -  b  -  2(c  -  d)    from     2(a  -  b)  -  c  +  d. 

Ans.  a  —  b  +  c  —  d. 

(5)  Subtract  a2  +  x2  -  ax  from  3a2  -  2ax  +  x2.  Ans.  2a2  -  ax. 

(6)  Multiply  2r3  +  4r2  +  8r  +  16  by  3r  -  6.  Ans.  6r4  -  96. 

(7)  Multiply  Am*  —  3/nn  —  n2  by  3/«  —  2n. 

Ans.  12m*  -  llmhi  +  3mn2  +  2n3. 

(8)  Divide  a6  -  6a4  +  a2  +  4  by  a2  -  1.  A«s.  a4  -  5a2  -  4. 

(9)  Divide  32s6  +  t*  by  2s  +  t.       Ans.  16s4  -  Ss3<  +  4s2t2  -  2st3  +  t*. 

(10)  Remove  the  signs  of  aggregation  from 

la  -  {3a  -  2[4a  -  3(5a  -  2)]}.  Ans.   12  -  18a. 

(11)  Remove  the  signs  of  aggregation  from 

15  -  2i/(x  +  y)  (x  -  2y)  -  4y[2x  -  3y(S  -  x)] 

Ans.   15  -  Sxy  +  36y2  -  10xy2  -  2x2y  +  <±y*. 

(12)  Enclose  within  parenthesis  the  third  and  fourth  terms,  also  the  fifth 
and  sixth  terms,  and  then  enclose  all  except  the  first  term  in  brackets,  of 
the  expression  1  -  20«  -  36mn  +  12n3  +  8?nhi  +  8n2. 

Ans.  1  -  4n[5  +  3(3m  -  n2)  -  2(m2  +  n)]. 

(13)  Change  the  signs  of  the  fraction  — -—ir  without  changing  its  value. 

c  -p  1 

,  7 

Ans. —  • 

c  +  1 

(14)  Change  the  signs  of  the  fraction     2  _  ^ — —j  without  changing  its 

,  ,  2a  -  8  2a  -  8 

value.  Ans.     —  - 


a2  -  2a  -  4       '  4  +  2a 


EQUATIONS 

24.  An  equation  is  an  expression  of  equality  between  two  sets  of 
quantities;  thus  43  =  64,  (a  +  6)2  =  a2  +  2ab  +  b-,  and  a2 
+  36  =  2a  —  7  are  equations.  It  will  be  noted  that  an  equation 
consists  of  two  parts,  or  members,  which  are  separated  by  the  sign 
of  equality.  The  part  on  the  left  of  the  sign  of  equality  is  called 
the  first  member,  and  the  part  on  the  right  is  called  the  second 
member.  In  the  first  two  equations  given  above,  the  second 
member  is  merely  another  way  of  writing  the  first  member,  the 
second  member  being  obtained  by  performing  the  operations 
indicated  in  the  first  member.  Equations  of  this  kind  are  not 
true  equations;  they  are  called  identical  equations.  The  third 
equation  is  a  true  equation,  and  since  the  second  member  cannot 
be  obtained  by  rewriting  or  performing  any- indicated  operations 
in  the  first  member,  it  is  called  an  independent  equation.  Every 
formula  such  as  was  given  in  Art.  1,  is  an  independent  equation. 


§2  MATHEMATICAL  FORMULAS  19 

It  is  to  be  noted  that  in  every  equation,  the  two  members  are 
exactly  equal,  in  the  same  sense  that  2  =  2  or  3  +  4  =  7. 
Bearing  this  in  mind,  the  following  law  will  at  once  be  evident: 

If  the  same  quantity  be  added  to  or  subtracted  from  both  members, 
if  both  members  be  multiplied  or  both  divided  by  the  same  quantity, 
or  if  both  members  be  raised  to  the  same  power  or  the  same  root  of 
both  members  be  taken,  the  equality  of  the  members  is  unaltered. 
For  instance,  take  the  identical  equation  16  =  16;  adding  4  to 
both  members,  16  +  4  =  16  +  4,  or  20  =  20;  if  4  be-subtracted 
from  both  members,  16  —  4  =  16  —  4,  or  12=  12;  if  both 
members  be  multiplied  by  4,  16  X  4  =  16  X  4,  or  64  =  64;  if 
both  members  be  divided  by  4,  16  -f-  4  =  16  -5-  4,  or  4  =  4;  if 
both  members  be  squared,  162  =  162,  or  256  =  256;  finally,  if 
the  square  root  of  both  members  be  taken  \/l6  =  \/l6,  or  4  =  4. 
What  is  true  of  an  identical  equation,  insofar  as  the  above  law  is 
concerned,  is  also  true  of  any  independent  equation.  For  ex- 
ample, if  x2  +  5x  =  16  and  4  be  subtracted  from  both  members, 
the  equation  becomes  x2  +  5a;  —  4  =  16  —  4  =  12;  and  if  16  be 
subtracted,  the  equation  becomes  x2  -\-  5x  —  16  =  0. 

25.  Transformation  of  Equations. — To  transform  an  equation 
is  to  make  alterations  in  the  members  without  destroying  their 
equality.     This  is  done  by  applying  the  law  of  Art.  24. 

Case  I. — A  term  may  be  transferred  from  one  member  to  the 
other  by  changing  its  sign;  this  is  called  transposition.  Consider 
the  equation  40  —  6a;  —16=  120  —  14a;.  Performing  the  opera- 
tions indicated  in  the  first  member,  24  —  6x  =  120  —  14a;;  this 
operation  is  called  collecting  terms.  Transposing  the  14.x  to  the 
first  member  and  the  24  to  the  second  member,  14a:  —  Qx  =  120 
—  24.  Collecting  terms,  8a:  =  96.  Dividing  both  members  by 
8,  x  =  12.  To  prove  that  x  =  12,  substitute  12  for  x  in  the 
original  equation;  if  the  two  members  are  then  equal,  8  is  the 
correct  value  for  x.  Thus,  40  -  6  X  12  -  16  =  120  -  14  X  12, 
or  40  -  72  -  16  =  120  -  168,  whence,  -  48  =  -  48,  and  the 
equation  is  said  to  be  satisfied. 

To  prove  the  rule  for  transposition,  consider  the  second  equa- 
tion above,  24  —  Qx  =  120  —  14a:.  If  14a;  be  added  to  both 
members  (which,  by  the  law  of  equations,  does  not  destroy  the 
equality),  the  equation  becomes  14a;  +  24  —  6a:  =  120  —  14a: 
+  14a;  =  120.  Note  that  as  the  result  of  this  operation,  the  term 
14a:  has  here  been  transferred  (transposed)  to  the  first  member 


20  ELEMENTARY  APPLIED  MATHEMATICS  §2 

and  that  its  sign  lias  been  changed  from  —  to  +  .  Subtracting 
l' 1  from  both  members  of  the  equation  14.r  -f  24  —  6x  =  120, 
the  equation  becomes,  14.r  -f  24  —  24  —  6x  =  120  —  24,  or 
1  l.r  —  6x.  =  96,  from  which,  Sx  =  96.  Note  that  as  the  result 
of  this  operation,  the  term  24  has  been  transposed  to  the 
second  member  and  that  its  sign  has  been  changed  from  -f  to  — . 
Evidently,  therefore,  a  term  may  be  transposed  from  one  member 
t<>  the  other  by  changing  its  sign.  Having  reduced  the  equation 
to  8x  =  96,  both  members  may  be  divided  by  8  without 
destroying  the  equality,  according  to  the  law  of  equations. 

XXX 

Case  II. — Find  the  value  of  x  in  ~  —  ^  -\-  _  =  44.     The  first 

£         o         O 

step  is  to  clear  the  equation  of  fraction,  which  may  be  done  by 
multiplying  both  members  by  each  denominator  in  succession. 

2x       2x 
Thus,  multiplying  first  by  2,  a; ~-  +  -=-  =  88;    multiplying 

6x 
by  3,  3x  —  2x  -f  -=r  =  264;  multiplying  by  5,  after  first  com- 
bining 3x  and   —  2x,  5x  -f  Qx  =  1320;  dividing  both  members 
by  11,  since  5x  +  6x  =  llz,  x  =  120.     Substituting  this  value 

f      .....     ,                     120       120   ,120        ..       . 
of  x  in  the  original  equation,  -= x-  -\ =-  =  44;  whence, 

60  —  40  +  24  =  44,  and  the  equation  is  satisfied. 

Multiplying  by  2,  3,  and  5  is  the  same  as  multiplying  by  2  X  3 

X  5  =  30.     Therefore,  instead  of  multiplying  by  these  numbers 

separately,  both  members  of  the  original  equation  might  have 

30  X  x 
been  multiplied  by  30,  and  the  result  would  have  been 

30  X  x  ,  30  X  x 
3 1 ^ —   =  30  X  44;  whence,  lox  -  lOx  +  6x  =  1320, 

1320 
or  llx  =  1320,  and  x  =  -^  =  120. 

Example.— Find  the  value  of  x  in  ^4^  +  %  =  4  -^~-  • 

2  3  4 

SOLUTION.— The  product  of  the  denominators  is  2  X  3  X  4  =  24.     Mul- 
tiplying both  members  by  24,  24(j"  +!>  +  2\x  =  9G  +  ^_^   since 

2  3  4 

x  —  5             5  —  x 
4        =   H r —     Since  all  the  numerators  are  now  divisible  by  the 

denominators,  perform  the  divisions,  thus  clearing  the  equation  of  fractions, 
and    obtain     12(x  +  3)  +  8x  =  96  +  6(5  -  x),    or     12j  4-  36  +  Sx  =  96 

+  30  —  6j.     Transposing  and  collecting    terms,  26x  =  90,  or  i  =  -~  = 

2b 
3ij.     Ans. 


§2  MATHEMATICAL  FORMULAS  21 

26.  The  subject  of  transformation  of  equations  is  of  great 

importance  in  connection  with  formulas.     For  instance,  consider 

the  formula, 

.37wT 
p  = 

v 

Suppose  that  it  is  desired  to  find  another  formula  giving  the 

value  of  iv.     Multiplying  both  members  of  this  equation  by  v,  pv 

=  .Z7wT,  which  may  be  written  .372V  =  pv.     Dividing  both 

members  by  .37 T, 

pv 
w  =  — - — > 
.37T 

a  formula  that  can  be  used  to  obtain  the  value  of  w. 

As  another  example,  transform  the  formula  t  =  .5SLJP  so  it 
can  be  used  to  find  the  value  of  P.  Dividing  both  members  by 
.53L,  the  equation  becomes  — -  =  J  ',  squaring  both  members 
p  £2  3.56£2  1 

=   5  =  ^809Z72  =  ~1J  '  since    ^809  =  3'5G  "'     Now 
multiplying  both  members  by  S, 

P  -  3.56iS<2 
L2 
A  numerical  factor  may  be  transferred  from  the  denominator 
to  the  numerator  by  multiplying  the  numerator  by  the  reciprocal 

of  the  factor.     Thus,  let  a  be  the  factor  and  --  the  fraction;  then 

an  ' 

M  -  X  m         -  X  m 

HL  _      a  a  a-  1 

an  "  1  =  ~rT'     Smce     ^809  =  3'56     very     nearl>'' 

-  X  a  Xn 
a 

t2  Z.56t2  t2  1  P  t* 

.2809L2  L2    "     °r'  .2809L2  ~   .2809  X  L2  ~  3"56  X  L2   = 

3Mt2 


L2 

27.  Prime  Marks  and  Subscripts^ — In  certain  formulas,  it  is 
desirable  to  use  the  same  letter  for  different  measurements  or 
quantities  of  the  same  kind.  The  formula  given  in  Art.  1  is  an 
instance.  Here  the  outside  diameter  was  represented  by  D  and 
the  inside  diameter  by  d.  Instead  of  using  a  capital  letter  for 
one  diameter  and  a  lower  case  letter  for  the  other,  the  outside 
diameter  might  have  been  represented  by  d'  and  the  inside 


ELEMENTARY  APPLIED  MATHEMATICS         §2 

diameter  by  d".  Had  there  been  Beveral  other  diameters  used  in 
the  formula,  they  might  be  indicated  by  d'",  d""  or  dlY,  dy ,  etc. 
These  marks  are  called  prime  marks,  and  they  are  read  in  connec- 
tion with  the  let  ins  as  f ollows :  d  prime,  d  second,  d  third,  d 
fourth,  etc.  In  connection  with  printing,  these  marks  are  called 
superior  characters. 

When  exponents  arc  used  in  connection  with  letters  affected  with 
prime  marks,  the  result  is  somewhat  awkward  (thus,  d'2,  d"2, 
d'"3,  etc.)  ami  it  is  then  customary  to  use  what  are  called  sub- 
scripts (in  printing,  called  inferior  characters),  which  are  small 
figures  of  the  same  size  as  those  used  for  exponents  and  placed 
slightly  below  and  to  the  right  of  the  letter;  for  example,  di 
(read  d  sub-one),  rf2  (read  d  sub-two),  d3  (read  d  sub-three),  etc. 
Exponents  can  be  used  with  such  letters  very  readily;  thus,  d\, 
d\,  dt.  etc.  Such  expressions  are  read  d  sub-one  square,  d  sub-tiro 
cube,  d  sub-three  fifth,  etc.  Always  remember  that  letters  having 
different  subscripts  or  prime  marks  are  just  as  different  and  are 
operated  on  in  the  same  way  as  though  different  letters  had  been 
used  instead.     For  instance,  bd'  +  8d"  -  2d'  =  3d'  +  8d". 

28.  Application  of  Formulas. — To  apply  a  formula,  substitute 
in  the  right-hand  member  the  values  of  the  letters  and  then  per- 
forin the  indicated  operations. 

Example   1. — Given  the  formula  I  =  lr2 T   +  on/^ — ZT  ^^  tne 

value  of  /  when  r  =  12.3  and  h  =  4.8. 

Solution. — Substituting  in  the  formula  the  values  of  r  and  h,  I  =  (12.31 

3    X   12.3X4.8    ,    3X4.82\        2x4.8 

~4~  +       20     j  3X12.3  -4.8  =  (15L29  ~  44'28  +  3456) 

X  .29907  =  110.466  X  .29907  =  33.037,   very  nearly.     Ans. 

Example  2.— Given  the  formula  S  =  h(A'  +  2^£1+_M^     Bnd  S 

4U'  +  VA'A"  +A') 
when  h  =  14.5,  A'  =  40.84,  and  A"  =  30.63. 

Solution. — Substituting  in  the  formula  the  values  given, 

S  =  14-5(40.84  +  2y/40.84  X  30.63  +  3  X  30.63) 
4(40.84  +  V40.84  X  30.63  +  30.63) 
_  14.5(40.84  +  70.737  +  91.89)       14.5  X  203.467 


4(40.84  +  35.3685  +  30.63)    '     4  X  106.8385 


=  6.9036 -•     Ans. 


Example  3.— Given  the  formula  p  =  „XT    ,    „,  to  find  D  when  p  =  56, 

ON  +  K  ^ 

P  =  200,  A*  =  700,  and  N   =  150. 

Solution.— First  transform  the  equation  so  D  will  stand  alone  in  the 

left-hand  member.     Multiplying  both  sides  of  the  equation  by  DN  +  K, 


§2  MATHEMATICAL  FORMULAS  23 

pND  +  pK  =  PK;  transposing  the  term  pK  to  the  .second  member,  pND 
=  PK  —  pK;  Dividing  both  members  by  pN , 

D  =PK-pK_K(P-p) 
p'N  pN 

Now  substituting  in  the  formula  for  D  the  values  of  K,  P,  p,  and  N, 
700(200-56) 
D  ~  ~ 56X"150~  =  12-     Ans- 


EXAMPLES 

(1)  Find  the  value  of  x  in  the  equation 

13  -  5(3  +  4x)  =  Ax  +  20  -  A(x  +  20).  Ans.  x  =  2.9. 

(2)  Find  the  value  of  x  in  the  equation 

x   .  2x   ,   7x       „/,        a;\ 

2  +T  +  8   =  7(1  ~  4)    +  35-  *»     *  =  nA- 

(3)  Find  the  value  of  x  in  the  equation 

15a  -  2ax  +  £   =  26  -  17a*.  Ans.  x  =  a(78  ~  45q). 

3a  1-|-  45aa 

(4)  When  w'  =  3,  w"  =  8.5,  u/"  =  3.6,  s'  =  .0951,  s"  =  1,  s'"  =  .1138 
t'  =  t"  =  60,  and  *'"  =  840,  find  the  value  of  t  in  the  formula. 

_  w's't'  +  iv"s"t"  +  w'"s'"t'" 
1  wVl^AHwV"  Ans-  *  -  94.753-. 

(5)  When  ff  -  178,  JT.  -  540,  and  Tc  =  1210,  find  p  from  the  formula 

u  Z7-6      7-9\ 
P  =  h\t    ~  ~fc)  v  =  L343- 

(6)  When  s  =  y2{a  +  b  +  c),  and  a  =  24.5,  6  =  37.8,  c  =  43.3,  find  r 
from  the  formula 

r  m  ^s-aHs-b)(B-e)m  Ans    r  m  g73M  + 

(7)  Given  the  formula  c  =  V2rh  ~-h*,  find  c  when  r  =  32 §  and  h  =  11  a. 

An*,     c  =  24.955  -. 

(8)  Given  n  =      .  find  n  when   L  =  1800,  w  =  3.1416,  d  =  8 

V7ra2  +  <2  '  » 

and   l  =  32'  *lns-     n  =  126.94  -. 


QUADRATIC  EQUATIONS 

29.  To  solve  an  equation  is  to  find  the  value  of  the  quantity- 
represented  by  some  particular  letter;  this  quantity  is  called  the 
unknown  quantity.  In  the  examples  given  in  Art.  25,  the  un- 
known quantity  is  x. 

What  is  termed  the  degree  of  an  equation  is  determined  by  the 
highest  exponent  of  the  unknown  quantity.  In  the  equations  so 
far  given,  the  highest  exponent  of  the  unknown  quantity  was  1; 


24  ELEMENTARY  APPLIED  MATHEMATICS  §2 

hence,  these  equations  arc  said  to  be  of  the  first  degree.  An 
equation  of  the  first  degree  is  also  called  a  linear  equation. 

When  the  highest  exponent  of  the  unknown  quantity  is  2,  as  in 
the  equation  Zx  +  9x2  -  12  =  7(2  +  x-),  which  reduces  to  2x2 
+  3.c  =  26  after  the  various  transformations  have  been  made, 
the  tiiuation  is  said  to  be  of  the  second  degree.  An  equation  of 
the  second  degree  is  generally  called  a  quadratic  equation. 

The  majority  of  equations  that  occur  in  practice  reduce  by 
transformation  to  linear  or  quadratic  equations.  Any  linear 
equation  may  be  represented  by 

ax  =  b,  (1) 

in  which  a  and  b  may  have  any  numerical  value,  and  may  be 
positive  or  negative.  If  a  be  negative,  it  may  be  rendered  posi- 
tive by  multiplying  both  sides  of  the  equation  by  —1  or,  what  is 
the  same  thing,  dividing  both  sides  (i.e.,  both  members)  by  —  1. 
Thus,  —  ax  —  b  divided  by  —1  becomes  ax  =  —b. 

Any  quadratic  equation  will  reduce  to  the  form 
ax2  +  bx  =  c,  (2) 

in  which  a,  b,  and  c  may  have  any  numerical  value,  and  may  be 
positive  or  negative.  If  a  be  negative,  it  may  be  rendered  posi- 
tive by  dividing  both  sides  of  the  equation  by  —  1.  Thus,  —ax2 
+  bx  =  —  c  divided  by  —  1  becomes  ax2  —  bx  =  c. 

In  equations  (1)  and  (2),  x  is  the  unknown  quantity  whose 
value  is  to  be  found,  and  which  may  be  represented  by  any  letter 
whatever  instead  of  x.  For  instance,  the  equation  might  be  n2-f- 
8n  =  33;  here  n  is  the  unknown  quantity,  a  =  1,  b  =  8,  and  c  = 
33.  In  the  equation  3s2  —  13s  =  — 14,  the  unknown  quantity 
is  s,  a  =  3,  b  =  — 13,  and  c  =  — 14. 

30.  It  is  sometimes  convenient  to  use  what  is  called  the  double 
sign  which  is  written  +  or  + .  The  first  character  is  read  plus 
or  minus,  and  the  second  character  is  read  minus  or  plus,  the 
name  of  the  upper  sign  being  pronounced  first,  both  characters 
being  a  combination  of  the  plus  and  minus  signs.  Either  charac- 
ter really  represents  two  signs  and  is  treated  as  two  signs,  whence 
the  name,  double  sign.  Thus,  16  ±  5  =  16  +  5  =  21  or  16  -  5 
=  11. 

Since  -\-a  X  +a  =  +  a2  and  -a  X  -a  =  -\-a2,  it  follows 
that  the  square  root  of  a°-  may  be  either  +a  or  —  a;  this  fact  is 
indicated  by  writing  \/a-  =  ±a,  consequently,  7  +  Vl6  =  11 
and  7  -  \/l6  =  3  may  be  represented  by  7  ±  -\/l6  =  11  or  3. 


§2  MATHEMATICAL  FORMULAS  25 

If  the  number  whose  square  root  is  to  be  extracted  is  negative, 
the  operation  can  only  be  indicated,  since  no  negative  number  can 
be  squared  that  will  give  a  positive^  product.  For  this  reason, 
such  expressions  as  V-  16,  V  -  c,  etc.  are  called  imaginary 
quantities.  Sincejhe  square  root  of  a  negative  quantity  is  in- 
dicated by  V-  a,  it  follows  that  the  square  of  an  imaginary 
quantity  is  a  negative  quantity;  thus,  (V^-  a)2  =  -  a,  (V -  16)2 
=  -  16,  etc. 

31.  Roots  of  Equations.— Any  value  of  the  unknown  quantity 
that  satisfies  the  equation  is  called  a  root  of  the  equation.  An 
equation  of  the  first  degree,  a  linear  equation,  has  but  one  root, 
while  an  equation  of  the  second  degree,  a  quadratic  equation, 

has  two  roots. 

To  solve  a  linear  equation,  reduce  it  to  the  form  ax  =  b;  then, 

dividing  both  members  by  a,  x  =  -»  and  the  result  thus  obtained 

is  the  root  of  the  equation. 

To  solve  a  quadratic  equation,  reduce  it  to  the  form 

ax2  -\-bx  =  c; 

then  substitute  the  values  of  a,  b,  and  c,  with  their  proper  signs, 

in  the  formula  

_  -  b  ±  Vb2  +  4ac 
x  2a 

This  formula  gives  two  values  for  the  unknown  quantity;  that  is, 
the  equation  has  two  roots. 

To  apply  the  formula,  let  the  equation  be  3s2  -  13s  -  - 14. 
Substituting  3  for  a,  - 13  for  b,  and  - 14  for  c, 

-r-13^W7r13)2+4X3X-  14 

s 2X3 

13  +  V/To9~=~T68  _  13  ±1  =  2i       2. 

= 6"  6 

If  either  of  these  values  be  substituted  for  8  in  the  original 
equation,  it  will  satisfy  the  equation.  Thus,  since  2|  =  i, 
oiv         ,yi=   41  _  -V-  =  -  V  =  -  14,  and  3  X  22  -  13 

«H"3V      —    loA33  3  -J  ' 

X  2  =  12  -  26  =  -  14-  Since  the  first  member  equals  the  second 
member  in  both  cases,  both  of  the  values  obtained  for  s  are  said 
to  satisfy  the  equation,  and  the  roots  of  the  equation  are  2£  and  2. 
The  above  formula  giving  the  value  of  x  is  so  important  that  it 
should  be  thoroughly  committed  to  memory. 


26  ELEMENTARY  APPLIED  MATHEMATICS  §2 

7 

Example  1. — Find  the  roots  of  the  equation  2x — =  =  5(21  —  x). 

x  -\~  o 

Solution. — Removing  the  parenthesis  and  transposing  to  the  first  mem- 
ber the  term  containing  x, 

7x ?-=  =  105 

x  +  5 

Clearing  of  fractions,  7x2  +  35x  -  7  =  105x  +  525 
Transposing  and  collecting  terms, 

7x2  -  70x  =  532 
Dividing  by  7,  x2  -  lOx  =  76 


-(-10)  +  Vl02  +  4  X  1  X76 

Substituting  in  the  formula,  x  = 

i£  X  1 

_  10  ±  ViOl        10  +  20.1- 

2  2 

=  15.05  or  —5.05.     Ans. 

7 
Substituting  15.05  for  x  in  the  original  equation,  2  X  15.05  —  tc  nr    i    g 

=  5(21-15.05).  Performing  the  operations  indicated,  29.75087+  = 
29.75,  which  agrees  close  enough  for  practical  purposes.     Substituting  —5.05 

7 
forx,  2  X  -5.05  -  _5  n5  .  5  =  5t21  _  (-5.05)].  Performing  the  opera- 
tions indicated,  129.9  =  130.25,  which  also  agrees  close  enough  for  most 
practical  purposes.  In  the  first  case,  the  difference  between  the  two  values 
was  00.00087,  and  in  the  second  case,  000.35.  If  more  accurate  results 
are  desired,  the  square  root  of  404  must  be  found  to  a  greater  number  of 
deicmal  places.  In  almost  every  case,  only  the  positive  root  is  desired, 
and  the  value  found  is  accurate  enough. 

x  4x2 

Example  2. — Find  the  value  of  x  in  3  —  4  —  x2  +  2x =-  =  45  — 

o  o 

3x2  +  4x. 

Solution. — Transposing  the  4  to  the  right-hand  member  and  the  two 

terms  containing  x  to  the  left-hand  member, 

*  +  2x2  -  2x  -  4|*  =  49  (1) 

Clearing  of  fractions, 

5x  +  30x2  -  30x  -  12x2  =  735 

Collecting  terms,  18x2  -  25x  =  735  

a  .    ...    .:      .     .,     ,         .  25  ±  V252  +  4  X  18  X  735 

Substituting  in  the  formula,  x  = 

Z  X  lo 

=  25  +  231.398- 

36 

=  7.1221+  or  -5.7333-.     Ans. 

Substituting  these  values  in  equation  (1),  which  is  practically  the  same 

7  1221  4  y7  12212 

as  the  original  equation,  ~ h  2  X  7.12212  -  2  X  7.1221 ^ 

=  49;  performing  the  indicated  operations,  48.999  =  49;  difference  =  00.001. 
Also,  -537333  +  2(-5.7333)2  -  2  X  -5.7333  -  4(~557333)'  =  49;  per- 
forming the  indicated  operations,  49.0004  =  49;  difference  =  00.0004. 
Both  results  are  close  enough  for  practical  purposes. 


§2  MATHEMATICAL  FORMULAS  27 

It  is  always  a  good  plan  to  substitute  at  least  one  of  the  roots 
for  the  unknown  quantity  in  the  original  equation  in  order  to 
make  sure  that  no  mistake  has  been  made  in  the  work. 


EXAMPLES 


Find  the  value  of  x  in  the  following  equations: 

(1)  x2  -  lGx  =  -63.  Ana.  x  =  9  or  7 

(2)  3x2  +  8x  =  5G.  Ana.  x   =  3.1882+  or  -5.8549- 

(3)  7x2  +  20x  =  32.  Ans.   i  =  l|or  -4 

(4)  x2  -  \x  =   If.  Ans.  x   =  If  or  -  j 

(5)  x3  +  (19  -  x)3  =   1843.  Ans.  x   =  11  or  8 

(6)  24.3x2  -  65.28x  =  427.486.  Ans.  x  =  5.74732  or  -3.0U0(J1. 


ACCURACY  IN  CALCULATION 

32.  Significant  Figures. — The  significant  figures  of  a  number 
are  those  figures  which  begin  with  the  first  digit  and  end  with  the 
last  digit  For  example,  in  the  numbers  2304000  and  .00023046, 
the  significant  figures  are  23046.  No  attention  is  paid  to  the 
decimal  point  in  connection  with  significant  figures,  and  ciphers 
to  the  left  of  the  first  digit  and  to  the  right  of  the  last  digit  are 
not  considered.  The  significant  part  of  a  number  is  that  part 
which  is  composed  of  its  significant  figures.  In  the  numbers 
.005236,  52.36,  and  5236000,  the  significant  part  is  5236,  and 
these  numbers  all  contain  four  significant  figures. 

In  practice,  before  any  formula  can  be  used,  it  is  usually  neces- 
sary to  make  one  or  more  measurements.  Thus,  before  the 
formula  in  Art.  1  can  be  used,  it  is  necessary  to  measure  the  length 
and  the  inside  and  outside  diameters  of  the  hollow  cylinder.  If, 
however,  it  is  assumed  that  the  inside  diameter  is,  say,  8  in.,  that 
the  weight  is,  say,  490  lb.,  and  it  is  desired  to  find  the  outside 
diameter,  it  may  be  readily  calculated  as  follows: 

W  =  .7854u>Z  (D  4-  d)  (D  -  d)  =  .7854  wl  (D2  -  d-),  whence, 

d  -  >/^+*°>Sx4^x3o+ 8!  - n-9943: 

in.  In  practice,  this  would  probably  be  taken  as  12  in.,  but  in 
any  case,  a  measurement  would  have  to  be  made  before  this 
result  could  be  applied. 

What  is  true  in  the  case  just  cited  is  also  true  in  practically 
every  case  that  occurs  in  the  application  of  formulas;  measure- 
ments of  some  kind — weight,  length,  area,  or  volume  must  be 
made  either  before  or  after  the  formula  is  used. 


28  ELEMENTARY  APPLIED  MATHEMATICS  §2 

33.  Accuracy  in  Measurements. — In  commercial  transactions 
and  in  ordinary  calculations  pertaining  to  engineering  matters, 
measurements  are  seldom  accurate  to  more  than  three  significant 
figures.  A  measurement  is  said  to  be  correct  to  n  significant  figures 
wlun,  if  expressed  to  n  figures,  the  figures  following  the  nth  figure 
being  considered  as  a  decimal,  the  difference  between  the  two 
numbers  is  greater  than  —.5  and  less  than  +.5.  Thus,  the 
number  3.141593  expressed  to  four  significant  figures  is  3.142; 
expressed  to  three  or  five  significant  figures  it  is  3.14  or  3.1416. 
Regarding  these  several  numbers  as  integers  and  subtracting  the 
original  number  from  them,  3142  —  3141.593  =  +.407,  which  is 
less  than  +.5;  314  -  314.1593  =  -.1593,  which  is  greater  than 
-.5;  and  31416  -  31415.93  =  +.07,  which  is  less  than  +.5. 
Hence,  3.141593  correct  to  3,  4,  or  5  significant  figures  is  3.14, 
3.142,  or  3.1416. 

When  a  number  is  expressed  to  a  certain  number  of  significant 
figures,  say  n,  and  it  is  necessary  to  employ  ciphers  to  make  up  the 
required  number  of  n  figures,  then  these  ciphers  are  counted  as 
significant  figures.  For  instance,  4599.996  and  4600.003  when 
expressed  to  6  significant  figures  are  written  4600.00. 

For  any  measurement  to  be  correct  to  n  significant  figures,  it 
is  necessary  to  know  that  the  figures  to  the  right  of  the  nth  figure 
form  a  number  that  is  greater  than  —.5  and  less  than  +.5. 
Suppose  that  4  tons  of  coal  (8000  lb.)  are  bought.  The  purchaser 
will  not  object  if  there  is  a  pound  less  and  the  dealer  wrill  not 
object  if  there  is  a  pound  more;  in  fact,  a  difference  of  5  or  possibly 
10  pounds  would  not  be  noticed.  To  obtain  4  significant  figures 
correct  in  this  transaction  would  require  that  the  amount  of  coal 
delivered  be  not  less  than  7999.5  lb.  and  not  more  than  8000.5  lb. 
But  the  consideration  of  half  a  pound  of  coal  would  be  ridiculous 
in  connection  with  a  weight  of  this  amount.  To  be  correct  to 
3  significant  figures,  the  coal  delivered  must  be  not  less  than  7995 
lb.  nor  more  than  8005  lb.,  and  if  the  amount  actually  delivered 
came  within  these  limits,  the  transaction  would  be  considered 
quite  accurate. 

What  is  true  of  a  heavy  weight  is  also  true  of  a  light  weight. 
Chemists  have  balances  so  accurate  that  when  the  proper  pre- 
cautions are  taken,  a  thousandth  of  a  grain  may  be  weighed;  but 
these  balances  arc  suitable  for  weighing  only  very  small  amounts, 
and  a  weight  of  even  a  tenth  of  a  pound  might  injure  them; 
possibly  a  hundredth  of  a  pound  would  be  the  greatest  amount  it 


§2 


MATHEMATICAL  FORMULAS 


29 


would  be  safe  to  use.  Since  a  pound  contains  7000  grains,  a 
hundredth  of  a  pound  contains  70  grains,  or  70,000  thousandths 
of  a  grain,  that  is,  5  significant  figures.  To  make  certain  that  the 
fifth  figure  was  correct,  it  would  be  necessary  that  the  balances 
detect  a  variation  of  half  a  thousandth  of  a  grain,  which  would  be 
beyond  the  limit  of  any  except  very  special  instruments ;  therefore, 
the  fifth  figure  could  not  ordinarily  be  determined,  and  the  weigh- 
ing would  be  correct  to  only  four  significant  figures. 

The  same  considerations  govern  any  other  kind  of  measure- 
ment, and  the  following  law  may  therefore  be  laid  down: 
measurements  may  be  considered  sufficiently  accurate  for  practical 
purposes  if  correct  to  3  significant  figures;  very  accurate,  if  correct  to 
4  significant  figures;  and  extremely  accurate,  necessitating  special 
methods  and  instruments,  if  correct  to  5  or  more  significant  figures, 

34.  Accuracy  in  Numerical  Operations. — Suppose  it  is  desired 
to  multiply  4343.944819  by  3.141593  and  obtain  the  result  correct 
to  5  significant  figures.  Using  all  the  figures  of  both  numbers, 
the  work  is  shown  at  (a),  and  the  product  to  5  significant  figures 


(a) 

(6) 

(c) 

4342.944819 

4342.94  + 

4'3'4'2.'9'4  + 

3.141593 

3.14159 
13028.82 

3.14159 

13028.834457 

13028.82 

434  2944819 

434  29 

4 

434  29 

173  71779276 

173  71 

76 

173  72 

4  342944819 

4  34 

294 

4  34 

2  1714724095 

2  17 

1470 

2  17 

39086503371 

39 

08646 

39 

13028834457 

13643.73 

68746 

13643.73 

13643.765042756667 

is  13644  —  .     Now  limiting  both  multiplicand  and  multiplier  to 

one  more  significant  figure  than  is  required  in  the  product,  in  this 
case,  5+1  =  6  figures,  the  work  is  shown  at  (6),  and  the  product 
to  5  significant  figures  is  13644  as  before.  Inspecting  the  work  at 
(b),  it  will  be  observed  that  if  a  line  be  drawn  as  shown  cutting  off 
all  figures  to  the  right  of  the  first  partial  product,  the  result  is 
13643.73,  which  becomes  13644  when  reduced  to  5  significant 
figures.  Observe  that  in  the  multiplication,  the  left-hand  digits 
of  the  multiplicand  and  multiplier  are  placed  under  each  other  and 
the  first  partial  product  is  found  by  multiplying  the  multiplicand 
by  the  first  figure  of  the  multiplier.  The  second  partial  product 
is  then  written  one  place  to  the  right  of  the  first  partial  product;  the 


30  ELEMENTARY  APPLIED  MATHEMATICS  §2 

third  partial  product  is  written  one  place  to  the  right  of  the 
second  partial  product;  etc.  This  brings  the  partial  products  in 
the  same  relative  positions  insofar  as  the  order  of  the  figures  of 
the  product  is  concerned  as  would  be  the  case  if  the  multiplying 
were  begun  wit  h  the  right-hand  figure  of  the  multiple. 

The  work  at  (c)  is  very  much  the  same  as  at  (6),  except  that  as 
soon  as  the  first  partial  product  has  been  obtained,  the  right- 
hand  figure  of  the  multiplicand  is  cut  off;  it  is  considered,  however, 
in  multiplying,  in  order  to  ascertain  how  much  to  carry.  In 
finding  the  second  partial  product,  there  is  nothing  to  carry  in 
this  case  because  1  X  4+  =  4+.  Having  found  the  second 
partial  product,  cut  off  another  figure  in  the  multiplicand;  then, 
since  4X9  =  36,  call  it  40  and  carry  4  to  the  next  product, 
obtaining  4X2  +  4  =  12;  write  2  and  carry  1.  Then,  4X4 
+  1  =  17;  write  7  and  carry  1.  4  X  3  +  1  =  13;  write  3  and 
carry  1.  4  X  4  +  1  =  17,  which  write.  Now  cut  off  another 
figure,  and  1  X  434  =  434,  which  write.  Cut  off  another  figure, 
and  since  5  X  4  =  20,  carry  2.  Then,  5  X  3  +  2  =  17;  write 
7  and  carry  1.  5X4+1  =  21,  which  write.  Finally,  cut  off 
one  more  figure,  and  9X3  =  27,  which  call  30  and  carry  3. 
Then  9  X  4  -f  3  =  39,  which  write.  It  will  be  observed  that 
the  first  figure  obtained  in  each  partial  product  is  written  directly 
under  the  last  figure  of  the  first  partial  product.  Adding  the 
partial  products  as  they  stand,  the  sum  is  13643.73;  or  13644 
when  reduced  to  5  significant  figures. 

Now  counting  the  number  of  figures  at  (a),  it  will  be  found  that 
109  figures  were  used;  at  (6),  64  figures  were  used;  at  (c),  44  figures 
were  used.  The  result  in  all  three  cases  is  the  same  when  the 
product  is  expressed  to  5  significant  figures.  Not  only  is  there  a 
great  saving  in  the  number  of  figures  employed,  but  there  is  very 
much  less  liability  of  making  mistakes,  and  it  is  very  much  easier 
to  check  the  work.  In  order,  however,  to  perform  the  work  as 
shown  at  (c),  it  is  necessary  to  begin  multiplying  with  the  first 
(left-hand)  figure  of  the  multiplier. 

As  another  example,  find  the  product  of  230.2585093  and 
15.70796  to  5  significant  figures.  The  work  is  shown  in  the 
margin.  Instead  of  reducing  the  numbers  to  5  -\-  1  =  6  signifi- 
cant figures,  they  are  written  as  they  stand  and  the  multiplicand 
is  limited  to  6  significant  figures  by  cutting  off  the  remaining 
figures.  The  first  partial  product  is  2302.59,  because  the  seventh 
figure  is  5+.     It  is  always  best  to  locate  the  decimal  point  in 


2'3'0 

.  '2'5'8'5093 

15.70796 

2302.59 

1151 

29 

161 

18 

1 

61 

21 

1 

3616 

.89 

§2  MATHEMATICAL  FORMULAS  31 

the  first  partial  product.  To  locate  it  in  this  case,  note  that  if  the 
multiplier  were  1.57  + ,  the  first  partial 
product  would  be  230.259;  but  the  multi- 
plier is  10  times  1.57+ ,  and  the  first 
partial  product  is  therefore  10  X  230.259 
=  2302.59.  After  multiplying  by  9  and 
obtaining  21  for  the  fifth  product,  cut  off 
one  more  figure  of  the  multiplicand,  and 
since  6  X  2  =  12,  write  1  for  the  sixth 
partial  product.  This  method  of  multi- 
plying is  slightly  more  accurate  than  that  at  (c)  in  the  preceding 
example,  when  both  factors  were  reduced  to  5  +  1  =  6  figures. 
The  product  correct  to  5  significant  figures  is  3616.9.  Ans. 
35.  Division  may  be  performed  in  a  somewhat  similar  manner. 

3616.890r2'3'0.'2'5'8'5093  TllUS'     to    find    the    quotient    of 

2302  59     15.70795-  361G-89    +   230.2585093     to     5 

significant  figures,  arrange  the 
work  as  shown  in  the  margin. 
Limit  the  divisor  to  5  +  1  =  6 
figures  by  cutting  off  all  figures 
beyond  the  sixth,  and  then  cut  off 
another  figure  in  the  divisor  every 
time  another  remainder  is  found. 
There  will  evidently  be  two 
integral  places  in  the  quotient, 
since  3616.890  -^  230.258+  = 
15.7+ .  The  quotient  correct  to  5  significant  figures  is  15.708. 
As  another  example,  find  the  quotient  of  13643.765  -5-  3.141593 

to  5  significant  figures.     Limiting 
13643.7'6'5(3/r4'lW3  ^    ^^    ^     ^^     to    6 

V2566_37       4342.94+  ^^  each>  thc  dividend  wiU  not 

contain  the  divisor;  hence,  instead 
of  cutting  off  another  figure  from 
the  divisor,  add  another  figure  to 
the  dividend,  and  then  proceed 
with  the  division  as  shown.  The 
quotient  evidently  contains  4  in- 
tegral places,  since  the  quotient  of 
13643  -5-  3  will  contain  4  integral 
places.  The  quotient  correct  to  5 
significant  figures  is  4342.9.     Ans. 


1314 

30 

1151 

20 

163 

01 

161 

18 

1 

83 

1 

61 

22 

21 

1 

1 

32  ELEMENTARY  APPLIED  MATHEMATICS  §2 

36.  Constants  and  Variables.— A  large  majority  of  the  formulas 
used  in  practice  contain  one  or  more  quantities  that  remain  the 
same  no  matter  what  the  conditions  are  that  govern  the  problem. 
For  instance,  the  formula  in  Art.  1  contains  the  quantity  .7854. 
Now,  no  matter  what  the  values  of  D,  d,  I,  and  w  are,  the  quantity 
.7854  remains  the  same;  for  this  reason,  it  is  called  a  constant. 
The  other  quantities  in  this  formula  are  called  variables,  because 
their  values  vary  or  change  for  different  cylinders.  For  all  cast- 
iron  cylinders,  w  =  .2604,  and,  hence,  for  all  cast-iron  cylinders, 
w  is  also  a  constant,  its  value  being  .2604. 

It  has  been  shown  that  for  multiplication  and  division,  the 
numbers  used  may  be  limited  to  one  more  figure  than  the  number 
of  significant  figures  required  in  the  result  without  loss  of  accuracy. 
What  is  true  in  this  respect  of  multiplication  and  division  is  also 
true  of  addition,  subtraction,  powers,  and  roots.  Consequently, 
in  all  practical  applications  of  formulas,  the  number  of  significant 
figures  used  in  all  the  quantities  contained  in  the  formulas  may  be 
limited  to  one  more  than  is  contained  in  the  constant  having  the 
smallest  number  of  significant  figures,  and  the  result  should  be 
limited  to  the  same  number  of  significant  figures  as  this  constant 

contains.     For   example,   the   formula  A  =  p~.    o  fi2   contains 

two  constants,  both  having  3  significant  figures;  hence,  the  values 
of  G  and  P  may  be  limited  to  3  +  1  =  4  significant  figures,  and 
the  result  when  found  should  be  expressed  to  3  significant  figures. 
The  formula  d  =  1.54aD  -+-  2.6  contains  two  constants,  one  of 
which  contains  but  two  significant  figures;  hence,  the  values  of  a 
and  D  may  be  limited  to  3  significant  figures  and  the  value  of  d 
when  found  should  be  limited  to  2  significant  figures. 

37.  In  the  examples  that  follow,  unless  for  some  reason  a 
special  exception  is  made,  all  applications  of  formulas  will  be 
made  under  the  assumption  that  all  the  quantities,  both  constants 
and  variables,  have  been  limited  to  one  more  significant  figure 
than  the  number  of  significant  figures  contained  in  the  constant 
having  the  least  number  of  significant  figures;  if  there  are  no 
constants,  then  all  quantities  should  ordinarily  be  limited  to  5 
significant  figures,  and  the  results  will  be  expressed  to  4  significant 
figures. 

WP 

Example. — In  the  formula  s  =  t^i  suppose  that  W  =  1200  pounds, 

I  =  108  inches,  E  =  1,500,000  pounds,  and  /  =  — -.  in  which  6  =  3  inches 
and  d  =  8  inches;  find  the  value  of  s. 


§2  MATHEMATICAL  FORMULAS  33 

Solution. — In  this  formula,  the  constants  48  and  12  are  exact,  and  are 
therefore  correct  to  any  number  of  significant  figures;  the  number  1,500,000 
is  correct  to  only  2  significant  figures,  and  it  is  doubtful  even  if  the  second 
figure  is  correct.  Consequently,  it  is  useless  to  employ  more  than  3  signifi- 
cant figures  in  the  calculation. 

Q    V  83 

First  calculate  the  value  of  J,  obtaining  I  =  =  128.     Substituting 

this  value  of  I  and  the  other  values  given  in  the  above  formula, 

9 
XW  V        27        27 

rm  x  io83      =  m  x  m  x  w  _  656i 
*?  x  xmm  x  128       mm  x  w    ~  40000  "  ,lb4' say  -16- 

4  15000  20000         22  Ans 

2 
The  formula  just  given  gives  the  value  of  s  in  inches,  s  being 
the  deflection  of  a  beam  having  a  certain  shape  and  under  certain 
conditions  of  loading,  etc.  Therefore,  the  deflection  in  this  case 
would  be  taken  as  .16  inch,  and  this  value  is  as  close  as  can  be 
obtained,  although  it  might  be  a  trifle  more  or  a  trifle  less.  Note 
that  instead  of  cubing  the  number  108,  the  cube  was  expressed  as 
the  product  of  three  factors,  in  order  to  employ  cancelation. 


APPROXIMATE  METHOD  FOR  FINDING  ROOTS 

38.  Cube  and  Fifth  Roots  of  Numbers. — In  certain  formulas 
used  in  engineering,  it  is  sometimes  necessary  to  extract  the  cube 
root  or,  less  frequently,  the  fifth  root  of  a  number.  These  roots 
may  be  found  by  an  extension  of  the  method  explained  in  Arith- 
metic for  finding  the  square  root;  but  it  entails  a  great  amount  of 
labor,  and  for  practical  purposes  an  approximate  method  answers 
the  purpose  equally  well  and  is  much  easier  to  apply.  Of  the 
many  approximate  methods  that  have  been  recommended,  the 
following  is,  perhaps,  the  simplest  and  most  accurate.  It  was 
discovered  by  Charles  Hutton,  a  famous  English  mathematician. 

Let  n  =  the  number  whose  root  is  to  be  found;  let  r  =  the 
index  of  the  root,  and  let  a  be  a  number  a  little  greater  or  a  little 
less  than  the  exact  value  of  the  root,  so  that  ar  is  a  little  greater 
or  a  little  less  than  n.     Then, 

(r  +  \)n  +  (r  -  \)ar 
.(r  -  l)rc  +  (r  +  1)< 
For  cube  root,  r  =  3,  and  formula  (1)  reduces  to 


r/~        \(r  +  \)n  +  (r  -  l)arl  .  ..,, 

Vw  =     7 7T ;     ,    /v  r  °>  nearly.  (1) 

L(r  —  l)n  +  (r  +  l)arJ 


+ 
3 


3/-        (2n  +  a\  ttv. 


34  ELEMENTARY  APPLIED  MATHEMATICS  §2 

For  fifth  root,  r  =  5,  and  formula  (1)  reduces  to 
5/-        /3n  +  2a5\  ,  . 

The  more  nearly  aT  approaches  in  value  to  n  the  more  accurate 
will  be  the  value  obtained  for  the  root.  In  order  to  save  labor  in 
finding  a,  the  following  table  has  been  calculated;  it  gives  the 
cubes  and  fifth  powers  of  1.1,  1.2,  1.3,  etc.  up  to  9.9,  and  is  used  as 
described  below. 

Suppose  it  is  desired  to  find  the  cube  root  of  34,586.  The  first 
step  is  to  point  off  the  number  into  periods  of  3  figures  each,  3 
being  the  index,  beginning  with  the  decimal  point  and  going  to 
the  right  and  left.  If  the  number  contains  an  integral  part, 
point  off  that  part  only;  but  if  it  is  a  pure  decimal,  point  it  off  to 
the  right.  Thus,  the  above  number,  when  pointed  off,  becomes 
34'586.  If  it  had  been  .034586,  it  would  become  .034'586  when 
pointed  off. 

The  next  step  is  to  move  the  decimal  point  so  that  it  will  follow 
the  first  period  that  contains  a  digit,  and  the  given  number  then 
becomes  34.586.  The  given  number,  after  shifting  the  decimal 
point  will  be  called  the  altered  number.  Of  course,  if  the  integral 
part  of  the  given  number  contains  not  more  than  3  figures,  it  is 
not  necessary  to  shift  the  decimal  point. 

Now  referring  to  the  table,  and  looking  in  the  column  headed 
n3,  the  number  34.586  is  found  to  He  between  32.768  =  3.23  and 
35.937  =  3.33;  the  cube  root  of  34.586  therefore  lies  between  3.2 
and  3.3,  and  one  of  these  two  numbers  is  to  be  selected  for  a  in 
applying  formula  (2).  To  decide  which  one,  find  which  of  the 
two  cubes  just  mentioned  is  nearest  in  value  to  the  altered 
number.  The  easiest  way  to  do  this  is  to 
add  the  two  numbers  and  divide  the  sum 
by  2  (this  is  the  arithmetical  mean  of  the 
two  numbers).  If  the  altered  number  is 
less  than  the  arithmetical  mean,  use  the 
smaller  number,  but  if  it  is  larger,  or  if  it 
is  near  the  arithmetical  mean  in  value,  use 
the  larger  number.  In  the  present  case, 
(32.768  +  35.937)  -s-  2  =  34.3525;  hence, 
use  the  larger  number.  Then  a  =  3.3  and 
a3  =  35.937.  The  work  of  applying  the 
formula  is  shown  in  the  margin.  After 
performing   the    division,  the  quotient  is 


34.586 

35.937 

2 

2 

69.172 

71.874 

35.937 

34.586 

105.109  ( 

106 . 460 

95  81  I 

.98731 

9  2950 

3  3 

8  5168 

2.96193 

7782 

296193 

7452 

3.258123 

3:^0 

319 

11 

§2 


MATHEMATICAL  FORMULAS 

CUBES  AND  FIFTH  POWERS 


35 


n 

»« 

B. 

n 

l»« 

n» 

1.0 

1.000 

1.00000 

5.5 

166.375 

5032.84375 

1.1 

1.331 

1.61051 

5.6 

175.616 

5507.31776 

1.2 

1.728 

2.48832 

5.7 

185.193 

6016.92057 

1.3 

2.197 

3.71293 

5.8 

195.112 

6563 . 56768 

1.4 

2.744 

5.37824 

5.9 

205 . 379 

7149.24299 

1.5 

3.375 

7.59375 

6.0 

216.000 

7776.00000 

1.6 

4.096 

10.48576 

6.1 

226.981 

8445.96301 

1.7 

4.913 

14.19857 

6.2 

238.328 

9161.32832 

1.8 

5.832 

18.89568 

6.3 

250.047 

9924 . 35643 

1.9 

6.859 

24 . 76099 

6.4 

262.144 

10737.41824 

2.0 

8.000 

32 . 00000 

6.5 

274 . 625 

11602.90625 

2.1 

9.261 

40.84101 

6.6 

287.496 

12523.32576 

2.2 

10.648 

51.53632 

6.7 

300 . 763 

13501.25107 

2.3 

12.167 

64.36343 

6.8 

314.432 

14539.33568 

2.4 

13.824 

79.62624 

6.9 

328.509 

15640.31349 

2.5 

15.625 

97.65625 

7.0 

343.000 

16807.00000 

2.6 

17.576 

118.81376 

7.1 

367.911 

18042.29351 

2.7 

19.683 

143.48907 

7.2 

373 . 248 

19349.17632 

2.8 

21.952 

172.10368 

7.3 

389.017 

20730.71593 

2.9 

24 . 389 

205.11149 

7.4 

405.224 

22190.06624 

3.0 

27.000 

243 . 00000 

7.5 

421.875 

23730.46875 

3.1 

29.791 

286.29151 

7.6 

438.976 

25355.25376 

3.2 

32 . 768 

335.54432 

7.7 

456.533 

27067.84157 

3.3 

35.937 

391.35393 

7.8 

474.552 

28871 . 74368 

3.4 

39 . 304 

454 . 35424 

7.9 

493.039 

30770 . 56399 

3.5 

42.875 

525.21875 

8.0 

512.000 

32768 . 00000 

3.6 

46.656 

604.66176 

8.1 

531.441 

34867.84401 

3.7 

50.653 

693.43957 

8.2 

551.368 

37073.98432 

3.8 

54 . 872 

792.35168 

8.3 

571.787 

39390.40643 

3.9 

59.319 

902.24199 

8.4 

592.704 

41821.19424 

4.0 

64 . 000 

1024.00000 

8.5 

614.125 

44370.53125 

4.1 

68.921 

1158.56201 

8.6 

636.056 

47042.70176 

4.2 

74 . 088 

1306.91232 

8.7 

658.503 

49842.09207 

4.3 

79.507 

1470.08443 

8.8 

681.472 

52773.19168 

4.4 

85.184 

1649.16224 

8.9 

704.9(i9 

55840.59449 

4.5 

91.125 

1845.28125 

9.0 

729 . 000 

59049.00000 

4.6 

97.336 

2059 . 62976 

9.1 

753.571 

62403.21451 

4.7 

103.823 

2293.45007 

9.2 

778.688 

65908.15232 

4.8 

110.592 

2548 . 03968 

9.3 

804.357 

60568.83603 

4.9 

117.649 

2824 . 75249 

9.4 

830.584 

73390.  in-'-' t 

5.0 

125.000 

3125.00000 

9.5 

857.375 

77378.09375 

5.1 

132.651 

3450.25251 

0.6 

884.736 

81537.26976 

5.2 

140.608 

3802.04032 

0.7 

912.673 

85873.40257 

5.3 

148.877 

4181.05403 

9.8 

941.192 

90392 . 07968 

5.4 

157.464 

4591.65024 

9.9 

970.299 

95099 . 00499 

5.5 

166.375 

5032.84375 

10.0 

1000.000 

100000.00000 

36  ELEMENTARY  APPLIED  MATHEMATICS  §2 

multiplied  by  a  =  3.3,  and  the  product  is  the  cube  root  of  34.586, 
approximately.  Since  the  decimal  point  was  shifted  one  period  to 
the  hfl  to  form  the  altered  number,  it  must  be  shifted  one  place 
to  the  right  in  the  root,  and  \/34,586  =  32.58123,  approximately. 
The  root  to  8  significant  figures  is  32.581178;  therefore,  the  root 
as  found  by  the  formula  was  correct  to  six  significant  figures. 

38.  The  root  when  calculated  as  just  described  will  always  be 
correct  to  at  least  five  significant  figures,  except  when  the  altered 
number  is  less  than  1.331  =  l.l3  and  is  nearly  equal  to  the  arith- 
metical mean  of  1  and  1.331.  This  case  will  be  discussed  in 
Art.  39. 

If  for  any  reason,  more  figures  are  desired  than  can  be  obtained 
by  proceeding  as  above,  express  the  root  as  found  to  3  or  4  signifi- 
cant figures,  and  substitute  it  for  a  in  this  formula.  Thus,  in  the 
preceding  example,  the  root  to  4  figures  is  32.58;  substituting  this 

•      ,         ,     /«x      3/ /2  X  34586  +  32.583\  no  ro 

for  a  in  formula  (2),   V  34,586  =  \34586  +  2  X  32.58V 

=  32.581177737944-  ;  the  root  correct  to  14  significant  figures  is 
32.581177737942— ;  hence,  the  root  as  calculated  was  correct  to 
13  significant  figures. 

The  reason  for  shifting  the  decimal  point  to  get  the  altered 
number  is  obvious — to  make  it  easier  to  use  the  table.  Since 
\/34586  =  v^lOOO  X  34.586  =  10-^34.586,  the  decimal  point 
must  be  moved  one  place  to  the  right  after  the  root  of  the  altered 

number    has    been    found.     Similarly,    v/. 034586  =  yj-  1fW| 

=  tV'V^34.586;  consequently,  if  the  decimal  point  is  shifted  one 
period  to  the  right  to  form  the  altered  number,  it  must  be  shifted 
one  place  to  the  left  in  the  root. 

39.  When  the  altered  number  is  less  than  1.2,  the  fifth  figure  of 
the  root  as  calculated  by  the  foregoing  method  will  usually  be 
incorrect.  In  such  cases,  proceed  as  follows:  Suppose  the  cube 
root  of  .001166  be  desired.  Pointing  off,  the  result  is  .001'166. 
Shifting  the  decimal  point  one  period  to  the  right,  the  altered 
number  is  1.166,  which  is  less  than  1.2.  Now  divide  the  decimal 
part  by  the  index  of  the  root,  and  the  quotient  is  .166  -f-  3  = 
.055+,  which  is  nearly  equal  to  the  decimal  part  of  the  cube  root 
of  1.166,  the  integral  part  being  1;  that  is,  \/lJ06  =  1.055, 
nearly.     Substituting  this  value  for  a  in  formula  (2), 

</Tm =  (i.^+lxS1-055  =  •997655025 x i-055 r 

1.052526051. 


MATHEMATICAL  FORMULAS 


37 


The   root   correct   to    12  significant  figures  is   1.05252604197. 
Hence,  \/.001166  =  .1052526+,  by  the  formula. 

40.  The  fifth  root  of  a  number  is  found  in  exactly  the  same 
manner,  using  formula  (3).  The  only  difference  in  the  process  is 
that  the  number  must  be  pointed  off  into  periods  of  five  figures 
each,  because  the  index  of  the  root  is  5.  As  an  example,  find  the 
fifth  root  of  214.83.  Here  the  integral  part  of  the  number 
contains  only  3  figures,  and  it  is  not  necessary  to  point  off  the 
number.  Referring  to  the  table,  the  given  number  falls  between 
2.95  =  205.11149  and  3.05  =  243;  the  arithmetical  mean  of 
these  two  numbers  is  224.055745,  and  as  this  is  greater  than 
the  given  number,  use  2.9  for  a  in  formula  (3).  Substituting 
in  the  formula, 

^2lT83  =  /3  X  214.88  +  2  X  205.11149V 

v  \2  X  214.83  +  3  X  205.11149/  AIMW/+. 

Ans. 
The  whole  calculation  is 
shown  in  the  margin.  It 
was  really  not  necessary  to 
use  so  many  decimal  places, 
but  since  but  very  little 
labor  would  be  saved  by 
using  a  smaller  number,  it 
was  not  considered  worth 
while  to  abbreviate  the 
process  further.  In  general, 
however,  only  one  more  signifi- 
cant figure  would  be  used  than 
was  desired  in  the  root. 


214.83  214.83 
3      2 


644.49  429.66 
205.11149   205.11149 
2         3 


410.22298 

615.33447 

644.49 

429.66 

1054.71298 

(1044.99447 

1044  99447 

1.00930006 

9  71851 

2.9 

9  40495 

2.01860012 

31356 

908370054 

31350 


2.926970174 


As  in  the  case  of  cube  root,  the  method  will  give  at  least  five 

significant  figures  correct,  except  when  the  altered  number  lies 

between  l5  =  1  and  l.l5  =  1.61051  and  is  near  the  arithmetical 

mean  of  these  two  numbers.     In  such  cases,  proceed  as  described 

in  connection  with  cube  root,  dividing  the  decimal  part  of  the 

altered  number  by  the  index,  in  this  case  5,  to  find  the  decimal 

part  of  the  root.     For  example,  find  the  fifth  root  of  1.308375. 

Here  .308375  -*-  5  =  .061675,  and   v/lT308375   =    1.06  nearly. 

Substituting  1.06  for  a  in  formula  (3), 

(3 X  1.308375+2 XI. 06*\  ,  __     ,  _„ 
2xT308375T3Xl.06'j  1^  =  1.055197-l  say  1.0552. 


•^1.308375=  ( 


Ans. 


38  ELEMENTARY  APPLIED  MATHEMATICS  §2 

The  root  correct  to  9  significant  figures  is  1.05522833.  If 
more  figures  of  the  root  are  desired,  substitute  1.055  for  a. 

41.  The  fourth  root  is  very  seldom  required.  In  case  it  should 
be  necessary  to  find  the  fourth  root,  all  that  need  be  done  is  to 
extract  the  square  root  and  then  extract  the  square  root  of  the 
result;  in  other  words,  \Yn  =  \\/n.  For  instance,  v/97.34  = 
\\  1)7.34  =  V9.8661  =  3.1410  +.  Here  the  square  root  of 
97.34  =  9.86610  + ,  and  the  square  root  of  9.8661  =  3.1410+ . 

42.  If  the  table  of  cubes  and  fifth  powers  is  not  available,  find 
the  first  figure  of  the  root  by  trial;  then  substitute  this  value  for 
a  in  formula  (2)  or  (3)  and  calculate  the  root  to  three  figures, 
calling  the  result  b.  Express  b  to  two  figures,  and  substitute  it 
for  a  in  formula  (2)  or  (3) ;  the  remainder  of  the  work  is  the  same 
as  before,  the  value  of  a  being  calculated  instead  of  being  taken 
from  the  table. 

Referring  to  the  example  of  Art  37,  the  altered  number  is 
34.586.     To  find  the  first  figure  of  the  root,  note  that  33  =  27 

27  +  64 
and  43  =  64;  the  mean  of  these  two  powers  is ~ =  45.5; 

hence,  use  3  for  the  first  figure  of  the  root.     Then,  a  =  3  and 
a3  =  33  =  27,  and 

'2  X  34.586  +  27\ 


3/Q,  ,0f.       V  X  34.o86  +  27\ 

V34-586 :  (34.086  +  2  x  27)  ■  3-2°- 


Expressing  this  root  to  two  figures,  b  =  3.3.     Substituting  this 
value  of  b  for  a.  in  the  formula,  3.33  =   35.937,  and 

*3/oTT^       /2  X  34.586  +  35.937\  _  _       _  0.010  . 
V34-586  =  134.586 +  2  X  35.937)  3-3  =  3-2o812  + 

It  will  be  observed  that  insofar  as  the  first  five  significant 
figures  are  concerned,  either  3.2  or  3.3  may  be  substituted  for 
a  in  the  formula;  but  3.3  is  a  slightly  better  value  in  this  case,  since 
it  gives  the  root  correct  to  6  figures,  while  3.2  gives  only  5 
figures  correct. 

The  same  procedure  will  be  followed  in  the  case  of  fifth  roots; 
thus,  referring  to  Art.  40,  the  altered  number  is  214.83.  Here 
35  =  243  and  25  =  32;  obviously,  3  is  the  proper  number  to  sub- 
stitute for  a  in  formula  (3),  and 

^21483  =  (Z  X  214'83  +  2  X  243\  ^       9QQ  i, 

V^4-83       \2  X  214.83  +  3  ^243/  3  =  2'93~  =  b- 

Expressing  b  to  two  figures,  the  root  is  2.9,  which  is  the  same  as 
the  value  of  a  used  in  Art.  40. 


§2  MATHEMATICAL  FORMULAS  39 

This  method  may  be  applied  to  the  special  cases  of  Arts.  39 
and  40.     Thus,  to  find  the  value  of  \^T7l66,  the  first  figure  of  the 

root  is  obviously  1,  and  (1  ' [iT^t  i) 1  =  L0524-  Now 
when  the  altered  number  does  not  differ  greatly  from  a3,  the 
value  of  b  may  be  calculated  to  4  or  5  significant  figures,  and  b 
may  be  expressed  to  3  or  even  4  figures,  if  desired,  before  sub- 
stituting for  a  in  the  formula;  but,  unless  great  accuracy  is 
desired,  it  is  best  to  express  b  to  3  figures,  to  save  labor  in  calcu- 
lation. In  the  present  case,  substitute  1.05  for  a  in  the  formula. 
Similarly,  in  Art.  40,  the  altered  number  is  1.308375,  and  the 
first  figure  of  the  root  is  1.     Hence, 

5A  QngQ7.  -    /3  X  1.308375  +  2X1\.        1  __,_ 
VIT08375  "  \2  X  1.308375  +Txl)  l  =  h°549- 
In  this  case,  1.05  or  1.055  may  be  substituted  for  a. 
If  desired,  this  method  may  be  used  for  finding  the  square 
root  of  numbers  instead  of  employing  the  exact  method  de- 
scribed  in  Arithmetic.     For  square   root,    r  =  2,   and   formula 
(1)  of  Art.  36  becomes 

/-       /3n  +  a2\ 
Vn  =  ln+^)  a 
As  an  example,  find  the  square  root  of  3.1416.     Here  a  is 
•j     xi     «         ,/3X  3.1416  +  4\ 
evidently  2,  and  (3  1416  +  3  x  4/  X      =  1-773+ .     Using  the 

c    4-^u       c  e  /n\„n      3  X  3.1416+ 1.772 

first  three  figures  for  a,  V3.1416  =  3  1416  +^  ^  1  772  X  L77  = 

1.77245594  —  .  By  the  exact  method,  the  root  to  9  significant 
figures  is  1.77245592;  hence,  the  root  as  found  by  the  formula 
was  correct  to  8  significant  figures. 

The  exact  method  is  so  easy  to  apply,  that  the  reader  is  ad- 
vised to  use  it  in  preference  to  the  formula. 


EXAMPLES 
Calculate  the  roots  in  the  following  examples  to  5  significant  figures: 

(1)  \Z.04608.  Arts.  .35851+. 

(2)  ^.86402.  Ans.  .97119+. 

(3)  v/1.0947.  Arts.  1.0306+. 

(4)  ^324,096,815.  Ans.  686.90-. 

(5)  ^4,063,972.  Ans.  20.979+. 

(6)  v/3.1416.  Ans.  1.4646 -. 


ELEMENTARY  APPLIED 
MATHEMATICS 

(PART  1) 


EXAMINATION  QUESTIONS 

(1)  Explain  the  difference  between  a  coefficient  and  an  ex- 
ponent,    (b)  What  is  meant  by  the  reciprocal  of  a  number  or 

2  c 

quantity?     (c)  What  is  the  reciprocal  of  a ?      Ajis.        _  ~- 

(2)  Add  a  —  b,b  —  c,  c  —  d,  and  d  —  c.  Let  a  =  20,  b  =  16, 
C  —  11,  and  d  =  8  and  prove  the  result  obtained  was  correct. 

Ans.  a  —  c. 

(3)  Find  the  sum  of:  a4  -  Sax2  -  Saz2  +  x2z2  +  z\  3ax2 
-  4a3  +  2xz2  +  6az2  -  x\  3a3  +  2a2z2  -  2x2z2  +  3a3z  -  5az2, 
4x4  -  6a2x  -  3a4  -  3z4  +  x2z2. 

Ans.  3x4  -  Qa2x  +  2xz2  +  3a3z  +  2a2z2  -  2az2  -  a3  -  2a4  -  2z4. 

(4)  Subtract  f a  +  36  +  \x  +  |<kc  from  4a  -  76  -  Sax  +  §a;. 

Ans.  a  -\-  \x  —  \ax  —  112. 

(5)  Multiply   a5  -  10a46  +  40a362  -  80a263  +  80a64  -  326s 
by  a  -  26. 

Ans.  a6  -  12a5b  +  60a462  -  160a363  +  240a2&4  -  192a65  +  6466. 

(6)  By  performing  the  indicated  operations,  reduce  the  follow- 
ing expression  to  a  simpler  form : 

a2(3a  -  5x)  -  (2a  -  x)2(a  -  2x)  +  (a2  +  ax  +  x2){a  -  x)  -  x\ 

Ans.  ax(7a  —  9x). 

(7)  Divide  z4  -  4x3  -  51a;2  -  Qx  +  120  by  x  +  5. 

Ans.  x3  —  9x2  —  6x  +  24. 
3  —  Hz 

(8)  Change  the  signs  of  the  fraction   5^.2  1^,9  without 

3  -  llx 
changing  the  value  of  the  fraction.  Ans.  9  _  6a.  _  5x2' 

(9)  Given  the  formula  v  =  v0  —  \at2,  rewrite  it  so  it  will  ex- 
press (a)  the  value  of  a,  and  (6)  the  value  of  t. 

2(t'0  -  v) 
a--     —p — 

Ans.  (  G77 c 

2(>o  -  v) 


\-4 


41 


42  ELEMENTARY  APPLIED  MATHEMATICS  §2 

(10)  Given  the  formula 


v  = 


1       ,00155 

n  s 


.5521  (23  +  m^) 


'Vs, 


.00155\n 
r 
find  the  value  of  v  when  n  =  .013,  s  =  .005,  and  r  =  .559. 

Ans.  v  =  3.6  +  . 
(11)  Remove  the  signs  of  aggregation  from 
(27  -  4z)[40  -  5(3*  +  7)]  +  6{24  -  (x  +  2)(x  -  4)[3 

-  5x(4  +  a?)]} 
Ans.  30a;4  +  60a:3  -  438a:2  -  1349a:  +  423. 

(11)  Divide    40x6   -   70a:5  +  4a:4  +  724a:3  +  88a;2  -  1080a: 
-  3390  by  5a;2  -  20a:  +  48. 

Ans.  8*'  +  18*  -  4x«  -  44,  -  120  -  £?£%*( 

(12)  Find  the  value  of  y  in  the  equation. 

3^  +  T  "  2  =  8  V9"  ~  y  +  3J 

Ans.  y  =  4.0223  + 

(13)  Find  the  two  values  of  x  in  the  equation  11a;2  —  35a:  =  40 

Ans.  x  =  4.0743+  or  -.89251- 

(14)  What  is  the  cube  root  of  705.33':  Ans.  8.9015  + 

(15)  What  is  the  fifth  root  of  .76054?  Ans.  .94673- 

(16)  What  is  the  value  of  V26.252  +  17.52?      Ans.  31.559+ 


ELEMENTARY  APPLIED 
MATHEMATICS 

(PART  II) 


MENSURATION  OF  PLANE  FIGURES 


DEFINITIONS 

43.  Mensuration  deals  with  the  measurement  of  the  length  of 
lines,  the  area  of  surfaces,  and  the  volume  of  solids;  its  principles 
and  rules  are  used  in  every  occupation  and  industry.  The 
subject  is,  therefore,  of  the  greatest  importance. 

Every  material  object  occupies  space  in  three  directions — in 
length,  in  breadth  or  width,  and  in  thickness  or  depth.  Every 
magnitude  or  body  is  consequently  said  to  have  three  dimensions 
— length,  breadth,  and  thickness. 

44.  A  mathematical  line  indicates  direction  only;  it  has  only 
one  dimension,  length,  and  has  no  breadth  or  thickness.  Such 
a  line  could  not  be  seen;  hence,  every  visible  line,  no  matter  how 
fine  it  may  be,  has  breadth,  but  when  considered  in  connection 
with  problems  relating  to  mensuration,  the  breadth  of  all  lines 
is  disregarded  and  they  are  conceived  as  having  only  length. 

45.  A  straight  line  is  one  that  extends  continuously  in  one 
unvarying  direction.     In  mathematics,  straight  lines  are  assumed 

to  be  infinite  in  length;  that  is       ^ 

they  are  assumed  to  be  capable  F 

of     being    extended    in    either 

direction  without  limit.  Fig.  3  shows  a  straight  line,  and  it  is 
assumed  that  it  can  be  extended  to  the  right  or  to  the  left  as  far 
as  is  desired — one  foot,  a  mile,  a  hundred  thousand  million  miles 
or  farther — without  any  change  in  its  direction.  In  practice, 
only  short  parts,  called  segments,  of  straight  lines  are  con- 
sidered; and  to  distinguish  one  line  from  another,  letters  or 
figures  are  placed  at  the  ends  of  the  segments.     Thus,  in  Fig.  3, 

43 


44 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


the  line  there  shown  is  called  the  line  AB,  when  it  is  considered 
as  extending  from  .4  to  B,  or  the  line  BA,  when  it  is  considered 
as  extending  from  B  to  A.  In  mathematics,  a  straight  line  is 
called  a  right  line,  and  will  usually  be  so  designated  hereafter. 

Another  definition  of  a  right  line  is :  a  right  line  is  the  shortest 
path  between  two  points.  This  definition  is  obvious,  since  if 
the  path  deviates,  even  in  the  slightest  degree,  from  a  point  to 
another  point,  the  path  (line)  will  be  longer  than  if  it  extended 
straight  from  one  point  to  the  other. 

46.  A  broken  line  is  one  that  is  made  up  of  two  or  more  right 
line  segments;  see  Fig.  4.  A  broken  line  is  distinguished  by 
placing  a  letter  (or  figure)  at  the  ends  of  each  segment.  Thus, 
the  broken  line  in  Fig.  4  is  called  the  line  abedef. 


Fig.  4. 

47.  A  curved  line  or  curve  is  one  that  has  no  right  line  seg- 
ment; its  direction  changes  continually  throughout  its  entire 
length.  See  Fig.  5.  A  curve  may  be  finite  (limited)  in  length 
or  infinite  (unlimited)  in  length,  but  in  either  case,  no  part  of  it 
is  ever  straight.  An  example  of  a  finite  curve  is  a  circle,  Fig. 
5,  and  an  example  of  an  infinite  curve  is  a  parabola,  see  (a), 


Fio.  6. 

Fig.  6.     When  only  a  portion  of  a  curve  is  considered,  it  is  called 
an  arc,  the  word  arc  meaning  bou\  alluding  to  its  shape. 

Curves  are  distinguished  by  placing  letters  at  the  ends  of  the 
arc  and  at  such  other  points  between  as  may  be  deemed  advisable. 
Thus,  in  Fig.  5,  the  circle  may  be  designated  as  ABCD;  in  (a), 


§2  MENSURATION  OF  PLANE  FIGURES  45 

Fig.  6,  the  curve  may  be  called  abc;  in  (6),  Fig.  6,  the  curve  may 
be  referred  to  as  12345. 

48.  When  two  lines  cross  or  cut  each  other,  they  are  said  to 
intersect,  and  the  place  where  they  intersect  is  called  the  point 
ot  intersection.  Thus,  in  Fig.  4,  b,  c,  d,  and  e  are  points  of  in- 
tersection of  the  lines  ab  and  cb,  of  be  and  dc,  etc.,  assuming  that 
these  segments  of  right  lines  are  prolonged  (produced);  in  Fig. 
5,  the  points  A,  B,  C,  and  D  are  the  points  of  intersection  of  the 
right  lines  AC  and  DB  with  the  circle. 

49.  In  mathematics,  a  point  indicates  position  only;  it  has  no 
dimensions.  In  practice,  a  point  resembles  a  very  small  circle — 
more  properly,  a  square — but  is  always  considered  to  be  without 
dimension.  A  point  may  always  be  conceived  as  being  the  point 
of  intersection  of  two  lines. 

50.  Parallel  lines  are  right  lines  that  are  always  the  same 

distance    from    each    other; c 

thev  never  intersect  no  mat-  „. 

^  ft — — — -D 

ter  to  what  length  they  may  7 

be  produced.     Thus,  in  Fig. 

7,  the  right  lines  AB  and  CD  are  parallel. 

51.  An  angle  is  the  difference  in  direction  between  two  right 
lines  that  intersect.     The  point  of  intersection  is   called  the 

vertex  of  the  angle,  and  the  lines  are 
called  the  sides.  Angles  are  designated 
by  placing  a  letter  at  the  vertex  and  by 
two  other  letters,  one  on  each  line.  In 
Fig.  8,  the  right  lines  AC  and  DB,  inter- 
secting in  the  point  0,  form  four  angles, 
designated  as  AOB,  BOC,  COD,  and 
DO  A.  Two  angles  on  the  same  side  of 
a  line  (which  forms  one  of  their  sides) 
and  separated  by  a  common  side  are  called  adjacent  angles. 
In  Fig.  8,  A  OB  and  AOD  are  adjacent  angles;  AOB  and  BOC 
are  also  adjacent  angles;  etc.  Note  that  adjacent  angles  have 
a  common  vertex,  a  common  side,  and  the  other  side  of  both 
angles  lies  in  the  same  right  line. 

52.  When  two  right  lines  intersect  in  such  a  manner  that  the 
adjacent  angles  are  equal,  the  lines  are  said  to  be  perpendicular 
to  each  other.     For  example,  in  Fig.  9,  CD  has  been  so  drawn 


46 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


/> 

Fig.  9. 


that  all  the  adjacent  angles.  COB  and  COA,  COB  and  BOD,  etc. 

are  equal;  hence,  AB  is  said  to  be  perpendicular  to  CD,  and  CD 
c  ie    -aid  to   be  perpendicular  to  AB.     All 

four  angles,  AOC,  COB,  BOD,  and  DO  A 
are  equal,  and  each  is  called  a  right  angle. 

_B  53.  A  horizontal  line  is  one  that  is 
parallel  to  the  horizon  or  to  the  water 
level;  with  the  book  held  in  its  usual 
position  for  reading,  AB,  Fig.  9,  is  a 
horizontal  line.  Any  line  that  is  perpen- 
dicular  to  a  horizontal  line  is  a  vertical 

line.     In  Fig.  9,  CD  is  a  vertical  line.     All  vertical  lines  have 

the  same  direction  as  a  phimb  line. 

54.  An  angle  that  is  smaller  than  a  right  angle  is  called  an 
acute  angle.  In  Fig.  10,  the  angle  AOB  is  evidently  smaller 
than  the  right  angle  COB;  hence  AOB  is  an  acute  angle. 

If  an  angle  is  greater  than 
a  right  angle,  it  is  called  an 
obtuse  angle.  In  Fig.  11. 
AOB  is  an  obtuse  angle, 
because  it  is  greater  than 
the  right  angle  COB.  In 
Fig.  8,  AOB  and  COD  are 
acute  angles,  and  AOD  and  BOC  are  obtuse  angles. 

55.  Angles  are  measured  in  several  ways,  the  most  common 
method  being  to  use  an  arc  of  a  circle.  This  method  is  called  the 
angular  measure  of  angles,  and  the  table  of  angular  measure  was 
given  in  Arithmetic.  The  entire  circle  is  supposed  to  be  divided 
into  360  equal  parts  called  degrees,  the  abbreviation  for  which  is 
(°) ;  each  degree  is  then  divided  into  60  equal  parts  called  minutes, 
the  abbreviation  for  which  is  (');  and  each  minute  is  divided  into 
60  equal  parts  called  seconds,  abbreviation  ("). 

The  draftsman  measures  an  angle  by  means  of  an  instrument, 
usually  made  of  metal  or  paper,  called  a  protractor,  which  is  a 
half  circle  divided  into  degrees  and  half  degrees  or  degrees  and 
quarter  degrees.  Evidently,  the  length  of  the  sides  of  an  angle 
have  nothing  to  do  with  the  size  of  the  angle,  since  lengthening 
or  shortening  the  sides  does  not  change  the  direction  of  the  lines. 
aequently,  by  laying  the  protractor  on  the  angle  in  such  a  man- 
ner that  the  center  of  the  half  circle  coincides  with  the  vertex  of 


Fig.    10. 


Fig.   11. 


§2 


MENSURATION  OF  PLANE  FIGURES 


47 


the  angle  and  the  bottom,  or  straight,  side  of  the  protractor 
coincides  with  one  side  of  the  angle,  the  other  side  will  cross  the 
arc  of  the  half  circle  and  the  size  of  the  angle  can  be  read  on  the 
graduated  edge.  This  is  clearly  shown  in  Fig.  12.  Here  the  pro- 
tractor P  is  so  placed  that  its  center  coincides  with  the  vertex  0 
of  the  angle,  and  the  bottom  side,  or  edge,  of  the  protractor  coin- 
cides with  the  side  OB  of  the  angle;  the  other  side,  OA,  of  the 


Fig.  12. 


angle  crosses  the  arc  at  59°,  which  is  the  size  of  the  angle.  More 
accurate  instruments  are  in  use,  but  they  all  are  based  on  the 
same  principle. 

55.  A  surface  has  no  thickness.  When  a  surface  is  perfectly 
flat,  so  that  a  right  line  may  be  drawn  on  it  in  any  direction  and 
anywhere  on  it,  the  surface  is  called  a  plane.  A  plane  surface 
may  be  tested  by  laying  a  straightedge  anywhere  on  it;  if  the 
straightedge  coincides  with  the  surface  throughout  the  length  of 
the  straightedge,  no  matter  where  it  is  placed,  the  surface  is  a 
plane  surface. 

Planes,  like  right  lines,  are  supposed  to  be  infinite  in  extent; 
that  is,  they  are  infinite  in  length  and  infinite  in  breadth;  but, 
like  lines,  only  parts  of  planes  are  considered  in  practice. 

56.  Two  planes  intersect  in  a  right  line.  For  example,  referring 
to  Fig.  13,  the  planes  ABCD  and  EFGH  intersect  in  the  right  line 
MN,  which  is  called  the  line  of  intersection. 

If  from  any  point  o  in  the  line  of  intersection  MN  a  line  oa 
be  drawn  in  the  plane  EFGH,  perpendicular  to  MN,  and  a  line 
ob  be  drawn  in  the  plane  ABCD,  also  perpendicular  to  MN, 
the  angle  aob  is  the  angle  which  the  planes  make  with  each  other. 


48         ELEMENTARY  APPLIED  MATHEMATICS  §2 

If  this  angle  is  a  right  angle,  in  which  case,  oa  and  ob  are  perpen- 
dicular to  each  other,  the  two  planes  are  then  said  to  be  perpen- 


Fig.  13. 


dicular  to  each  other;  otherwise,  they  make  an  acute  or  obtuse 
angle  with  each  other,  according  as  the  angle  aob  is  acute  or 
obtuse. 


PLANE  FIGURES 

57.  A  plane  figure  is  any  outline  that  can  be  drawn  on  a  plane 
surface;  and,  since  a  plane  is  infinite  in  length  and  breadth,  a 
plane  figure  may  be  of  any  size  and  any  shape.  To  be  complete, 
however,  the  figure  must  close;  that  is,  if  the  figure  be  traced 
by  moving  the  pencil  over  it,  then,  starting  from  any  point  on 
the  outline  and  passing  over  the  entire  figure,  the  pencil  must 
return  to  the  point  from  which  it  started. 

58.  Polygons. — The  simplest  plane  figures  are  those  made  up 
entirely  of  right  lines — called  polygons,  which  means  many  angles, 
from  poly  (meaning  many)  and  gonia  (meaning  angles),  and  so 
called  because  a  polygon  always  has  as  many  angles  as  it  has  sides. 

The  right  lines  i  hat  form  the  outline  of  a  polygon  are  called  the 
sides  of  the  polygon,  and  the  angles  included  between  the  sides 
are  called  the  angles  of  the  polygon.  The  sum  of  the  lengths  of 
the  sides  is  called  the  perimeter  of  the  polygon.  The  perimeter, 
therefore,  is  the  length  of  the  outline  that  bounds  the  polygon; 
it  equals  the  distance  around  it. 


§2 


MENSURATION  OF  PLANE  FIGURES 


49 


59.  Polygons  are  named  in  accordance  with  the  number  of 
sides  that  they  have.  A  polygon  with  three  sides  is  called  a 
triangle;  one  with  four  sides  is  called  a  quadrilaterial;  one  of  five 
sides  is  a  pentagon;  one  of  six  sides  is  a  hexagon;  one  of  seven  sides 
is  a  heptagon;  one  of  eight  sides  is  an  octagon;  one  of  nine  sides 
is  a  nonegon;  one  of  ten  sides  is  a  decagon;  one  of  eleven  sides 
is  an  undecagon;  one  of  twelve  sides  is  a  dodecagon;  etc.  In 
practical  work,  the  triangle,  quadrilateral,  and  hexagon  are  very 
freely  used,  and  the  pentagon,  octagon,  and  decagon  are  oc- 
casionally used;  the  other  polygons  are  practically  never  used. 

60.  A  regular  polygon  is  one  in  which  all  the  sides  and  all  the 
angles  are  equal.     Fig.  14  shows  regular  polygons  of  3,  4,  5,  6,  8, 


B 


C   A 


n         a 


Triangle  Quadrilateral 

O, vD 


Pentagon 


Hexagon 


Octagon 
Fig.  14. 


Decagon 


and  10  sides.  In  each  figure,  the  sides  are  all  equal  in  length,  and 
the  angles  (called  the  interior  angles)  are  all  equal.  Thus,  in 
the  regular  pentagon,  AB  =  BC  =  CD  =  DE'=  EA,  and  ABC 
=  BCD  =  CDE  =  DEA  =  EAB. 

61.  To  find  the  number  of  degrees  in  one  of  the  equal  angles 
of  a  regular  polygon,  let  n  —  the  number  of  sides  and  a°  =the 
number  of  degrees  in  one  of  the  equal  angles;  then 

a°  =  (n  ~  2)  X  180° 
n 

For  instance,  in  the  regular  triangle,  the  interior  angles  are 

3—2 
equal  to  — 5 — ■  X  180°  =  60°;    in    the    regular   quadrilateral, 

o 

a0  =      ~      X  1S0°  =  90°;  in  the  regular  hexagon,  a°  =  — ^ — 

X  180°  -  120°;  etc. 
4 


50 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


TRIANGLES 

62.  Triangles  are  classified  in  two  ways:  according  to  their 
angles,  and  according  to  their  sides.  When  the  angles  are  con- 
sidered, triangles  are  called  right  triangles,  when  they  have  one 
right  angle;  oblique  triangles,  when  they  have  no 
light  angle;  and  equiangular  triangles,  when  all  the 
angles  are  equal.  An  equiangular  triangle  is  also 
an  oblique  triangle.  Fig.  15  shows  a  right  triangle, 
the  right  angle  being  at  B.  It  may  here  be  re- 
marked that  when  there  is  no  possibility  of  misun- 
derstanding, angles  may  be  named  by  a  single 
letter  placed  at  the  vertex.  Thus,  in  Fig.  15, 
angle  C  is  the  angle  ACB,  angle  A  is  the  angle  CAB,  and  the 
angle  B  is  the  angle  ABC. 

When  the  sides  are  considered,  triangles  are  called  isosceles, 
scalene,  or  equilateral  triangles.  An  isosceles  triangle  is  one  that 
has  two  equal  sides,  see  Fig.  16;  here  CA  =  CB.  A  scalene 
triangle  is  one  in  which  all  the  sides  have  different  lengths,  see 
Fig.  17.  An  equilateral  triangle  is  one  in  which  all  the  sides  are 
equal,  see  Fig.  18.  A  scalene  triangle  is  always  an  oblique  tri- 
angle.    An  equilateral  triangle  is  also  an  isosceles  triangle;  it 


Fig.  15. 


Fig.   16. 


Fig.   17. 


Fig.  18. 


is  likewise  an  equiangular  triangle.  A  right  triangle  is  usually 
a  scalene  triangle,  but  if  the  two  sides  that  enclose  or  form  the 
right  angle  are  equal,  it  is  then  an  isosceles  triangle  also. 

For  convenience  in  mathematical  operations,  triangles  are 
lettered  with  capital  letters  at  the  vertexes  and  with  small  letters 
(lower  case  letters)  placed  at  the  middle  of  the  sides,  the  lower 
case  letters  being  in  each  instance  the  same  as  the  capital  letters 
at  the  angles  opposite  the  sides.  Thus,  referring  to  Figs.  15-18, 
side  a  is  opposite  angle  A,  side  b  is  opposite  angle  B,  and  side  c 
is  opposite  angle  C. 


§2  MENSURATION  OF  PLANE  FIGURES  51 

The  side  on  which  any  triangle  (or  any  polygon)  is  considered 
as  standing,  the  plane  of  the  figure  being  supposed  to  be  vertical, 
is  called  the  base.  In  Fig.  14,  AC  is  the  base  of  the  triangle; 
AD  is  the  base  of  the  quadrilateral;  and  AE  is  the  base  of  the 
pentagon.  In  Fig.  15,  CB  is  the  base;  in  Figs.  16-18,  A B  is  the 
base. 

63.  Some  Properties  of  Triangles. — (1)  In  any  triangle,  the 
sum  of  the  three  angles  is  always  180°;  thus,  in  Figs.  15-18, 
A  +  B  +  C  =  180°.  Therefore,  if  two  of  the  angles  arc  known, 
the  third  can  be  found  by  subtracting  the  sum  of  the  two  known 
angles  from  180°.  Thus,  in  Fig.  17,  if  A  =  68°  23'  and  B  =24° 
35',  C  =  180°— (68°  23'  +  24°  350  =  87°  2'. 

(2)  In  a  right  triangle,  one  of  the  angles  being  a  right  angle 
and  therefore  equal  to  90°,  the  sum  of  the  other  two  angles  must 
be  180°  —  90°  =  90°;  consequently,  both  angles  must  be  acute, 
since  neither  can  exceed  (nor  even  equal)  90°.  Also,  this  sum 
must  be  90°;  because  the  sum  of  the  three  angles  is  180°,  and 
since  one  of  the  angles  if  90°,  the  sum  of  the  other  two  angles 
must  be  180°  -  90°  =  90°.  Hence,  if  one  of  the  acute  angles  of 
a  right  triangle  is  known,  the  other  can  be  found  by  subtracting 
the  known  angle  from  90°.  Thus,  in  Fig.  15,  suppose  that  the 
angle  A  =  28°  42';  then  the  angle  C    =  90°  —28°  42'  =  61°  18'. 

(3)  The  longest  side  is  always  opposite  the  greatest  angle,  and 
the  greatest  angle  is  always  opposite  the  longest  side.  In  Figs. 
15  and  17,  b  and  c  are  respectively  the  longest 
sides;  hence,  B  and  C  are  respectively  the 
largest  angles. 

(4)  If  any  two  sides  of  a  triangle  are  equal, 
the  angles  opposite  those  sides  are  also  equal. 
In  Fig.  16,  a  and  b  are  equal;  hence  A  =  B. 
Consequently,  in  every  isosceles  triangle,  two 
of  the  angles  are  equal. 

(5)  If  an  isosceles  triangle  be  so  placed 
that  the  unequal  side  forms  the  base,  as  in 
Fig.  19,  and  a  perpendicular  to  the  base  is  drawn  from  the 
vertex  of  the  angle  opposite,  the  perpendicular  divides  the  base 
into  two  equal  parts.  The  perpendicular  is  then  said  to  bisect  the 
base,  the  word  bisect  meaning  to  cut  in  halves.  In  Fig.  19,  AC  = 
CB,  and  CD  is  perpendicular  to  AB;  therefore,  AD  =  DB. 

(6)  If  two  angles  of  one  triangle  are  equal  to  two  angles  of 
another  triangle,  the  third  angle  of  the  first  triangle  is  equal  to 


52 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


the  third  angle  of  the  second,  because  the  sum  of  the  two  angles 
subtracted  from  180°  is  the  same  in  both  cases.  In  Fig.  20,  if 
A  =  A'  and  C  =  C,  then  B  =  B'. 

64.  Similar  Triangles. — If  the  angles  of  one  triangle  are  equal 
to  the  angles  of  another,  the  two  triangles  are  said  to  be  similar. 
If  the  sides  of  one  triangle  are  equal  to  the  sides  of  another,  then 
the  triangles  are  equal. 

When  two  triangles  arc  similar,  as  ABC  and  A'B'C  in  Fig.  20, 
one  may  be  superposed  on  the  other;  that  is,  the  vertex  of  one  of 


Fig.  20. 


the  angles  of  one  triangle  may  be  placed  over  the  vertex  of  the 
equ;il  angle  of  the  other  triangle,  and  the  sides  of  the  two  angles 
can  then  be  made  to  coincide,  since  they  both  have  the  same  direc- 
tion from  the  common  vertex,  see  Fig.  21.  Here  the  vertex 
C  of  the  angle  C  in  Fig.  20  is  placed  over  C,  and  the  sides  CA 
and  CB  are  made  to  coincide  with  the  sides  CA'  and  C'B';  then 
AB  is  parallel  to  A'B'. 

65.  When  two  triangles  are  similar,  the  sides  opposite  the  equal 
angles  arc  proportional.  Thus,  in  Fig.  20,  if  the  triangles  ABC 
and  A'B'C  are  similar,  and  angle  A  =  angle  A'  and  angle  C 
=  angle  C,  then  AB  :  A'B'  =  AC  :  A'C;  also,  AB  :  A'B'  =  BC: 
B'C,  and  BC  :  B'C  =  AC  :  A'C.  This  is  a  very  important 
principle,  and  should  be  remembered. 

Had  the  angle  A  been  superposed  on  the  angle  A',  Fig.  20,  the 
result  would  be  as  shown  in  Fig.  21,  the  dotted  line  BC  being  paral- 
lel to  B'C.  That  AB,  Fig.  21,  is  parallel  to  A'B'  is  evident  from  the 
fact  that  since  angle  A  =  angle  A'  and  the  sides  AC  and  A'C 
coincide,  AB  has  the  same  direction  as  A'B' ,  and  when  two  right 


§2  MENSURATION  OF  PLANE  FIGURES  53 

lines  have  the  same  direction,  they  must  either  coincide  or  be 
parallel.  For  the  same  reason,  BC  is  parallel  to  B'C.  Hence, 
if  two  sides  and  an  angle  of  one  triangle  are  equal  to  two  sides  and 
an  angle  of  another  triangle,  the  triangles  are  equal. 

66.  The  Right  Triangle. — The  right  triangle  is  so  important 
that  it  requires  special  treatment. 
Referring  to  Fig.  22,  ABC  is  a  right 
triangle,  right-angled  at  C.  The 
side  C,  which  is  opposite  the  right 
angle,  is  called  the  hypotenuse;  the 
hypotenuse  is  alwaj's  the  longest  side 
of  a  right  triangle,  since  it  is  opposite 
the  largest  angle.  The  other  two 
sides  are  called  the  short  sides  or 
legs.  In  every  right  triangle,  the  square  of  the  hypotenuse  is 
equal  to  the  sum  of  the  squares  of  the  legs;  that  is,  referring  to 
Fig.  22, 

C2    =    a2  _|_   tf       (J) 

This  is  a  very  important  principle,  and  should  be  memorized; 
it  is  used  more  than  any  other  principle  in  mensuration.  By 
means  of  it,  any  side  of  a  right  triangle  can  be  found  if  the  lengths 
of  the  other  two  are  known.     For  example,  if  a  and  b  are  known, 


c  =  Va2  +  b2     (2) 

If  c  and  a  are  known, 

b  =  Vc2  -  a2     (3) 

If  c  and  b  are  known, 

a  =  Vc2  -  62     (4) 

Formulas  (2),  (3),  and  (4)  are  derived  from  formula  (1)  by 
solving  (1)  for  c,  b,  and  a. 

Example  1. — Referring  to  Fig.  22,  suppose  the  length  of  AC  is  \%  in. 
and  the  length  of  BC  is  2%  in.;  what  is  the  length  of  AB? 

Solution. — Since  4%  =  4.375,  and  2%  =  2.875,  substitute  the  values 
for  a  and  b  in  formula  (2),  obtaining 

c  =  V4.3752  +  2.875*  =  5.2351  -  in.     Ans. 

It  is  evident  that  formula  (1)  might  have  been  used  instead  of 
formula  (2);  in  fact  formula  (1)  may  be  used  in  every  case  of  this 
kind. 

Example  2. — In  Fig.  23,  P  and  P'  are  two  pulleys,  whoso  centers  are  10 
ft.  8J>2  in.  and  4  ft.  6%  in.,  respectively,  from  the  floor  AB,  which  is  sup- 
posed to  be  level.  By  dropping  plumb  lines  from  the  shafts,  the  horizontal 
distance  between  the  pulley  centers  is  found  to  be  14  ft.  7L4  in.  What  is 
the  distance  O'O  between  the  pulley  centers? 


54 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


Solution. — Since  AB  is  horizontal  and  OC  and  O'D  are  plumb  lines,  the 
angle  OCD  is  a  right  angle.  Drawing  O'E  parallel  to  OCD,  it  must  be  per- 
pendicular to  OC  (since  AB  is  perpendicular  to  OC),  and  O'EO  is  also  a  right 
angle.  Drawing  O'O,  O'EO  is  a  right  triangle,  in  which  the  side  O'E  =  14 
ft.  7.25  in.  and  OE  =  10  ft.  8.5  -  4  ft.  6.75  in.  =  6  ft.  1.75  in.     For  con- 


Fig.  23. 

venience  in  calculation,  reduce  the  lengths  of  these  sides  to  inches,  obtain- 
ing O'E  =  175.25  in.  and  OE  =  73.75  in.  Substituting  these  values  in 
formula  (2),  O'O  =  Vl75.252  +  73.752  =  190.14-  in.  =  15  ft.  10M  in., 
very  nearly  (0.14  =  )4,  nearly).     Ans. 

67.  Equality  of  Angles. — In  connection  with  the  last  example, 
it  was  stated  that  the  angle  O'EO  was  a  right  angle,  because 
O'E  was  perpendicular  to  OC.  It  might  have  been  stated  that 
O'EO  is  a  right  angle  because  it  is  equal  to  OCD,  which  is  a  right 


Fiq.  24. 

angle,  in  accordance  with  the  following  principle:  If  the  two 
sides  of  one  angle  are  parallel  to  the  two  sides  of  another  angle, 
both  sides  of  which  extend  in  the  same  or  both  in  opposite 
directions,  the  two  angles  are  equal.  Thus,  referring  to  Fig.  24,  if 
A'O'  be  parallel  to  AO  and  B'O'  be  parallel  to  BO,  both  sides 
extending  in  the  same  direction,  A'O'B'  is  equal  to  AOB.  Again, 
if  A"0"  be  parallel  to  AO  and  B"0"  be  parallel  to  BO,  both  sides 


§2 


MENSURATION  OF  PLANE  FIGURES 


55 


extending  in  opposite  directions  to  AO  and  BO,  A"0"B"  is  equal 
to  AOB.  If  one  side  of  an  angle  coincide  with  a  side  of  another 
angle,  and  the  second  side  of  the  first  angle  be  parallel  to  the  second 
side  of  the  other  angle,  the  two  angles  are  equal.  Thus,  if  0'" 
B'"  and  0""B""  be  parallel  to  OB,  Fig.  24,  AO'"B"'  and  OQ'" 
B'"  are  equal  to  AOB.  This  a  special  case  of  the  above  principle, 
in  which  two  of  the  parallel  lines  coincide.  For  the  same  reason 
AvOBv  =  00"" B""  =  AOB.  Referring  to  Fig.  23,  OE  coincides 
with  OC,  and  since  O'E  is  parallel  to  CD,  O'EO  =  DCO. 


Fig.  25. 

A  second  principle  applying  to  equality  of  angles  is:  If  the 
two  sides  of  one  angle  are  perpendicular  to  the  two  sides  of  another 
angle  and  both  angles  are  acute  or  both  obtuse,  the  two  angles  are 
equal.  Referring  to  Fig.  25,  suppose  that  A'O'  and  B'O'  are  per- 
pendicular to  AO  and  BO,  respectively;  then  A'O'B'  =  AOB. 
Again,  suppose  that  0"B"  and  0"A"  are  perpendicular  to  OB 
and  OA,  respectively;  then  B"0"A"  =  BOA.  Finally,  suppose 
that  A'"0'"  and  B'"0'"  are  respectively  perpendicular  to  AO 
and  BO  produced  (indicated  by  the  dotted  lines;  then,  by  the 
first  principle,  A"" OB""  =  AOB,  and  A'"0"'B"'  =  AOB. 

These  two  principles  are  very  important  and  are  frequently 
used  in  connection  with  practical  problems. 

68.  Area  of  Triangles. — In  Fig.  26,  let  AC  be  the  base  of 
the  triangle  ABC.  From  the  vertex  of  the  angle  B  opposite  the 
base,  drop  a  perpendicular  BD;  the  point  where  the  perpendicular 
cuts  the  base  is  called  the  foot  of  the  perpendicular,  and  that  part 


56 


ELEMENTARY  APPLIED  .MATHEMATICS 


§2 


of  it  included  between  the  vertex  B  and  the  foot  D  is  called  the 
altitude  of  the  triangle.  The  word  altitude  means  height,  and 
it  is  usually  denoted  in  formulas  by  the  letter  h.  In  Fig.  27, 
it  is  necessary  to  produce  the  base  in  order  that  it  may  be  cut  by 
a  perpendicular  from  B.     In  both  figures,  h  =  BD  is  the  altitude. 


C  A 


In  any  triangle,  the  area  is  equal  to  half  the  product  of  the  base 
and  altitude.  Hence,  letting  A  =  the  area,  b  =  the  base,  and 
h  =  the  altitude, 

A  =  \bh  =  — £— 

Any  side  may  be  taken  as  the  base,  the  altitude  being  the 
perpendicular  distance  between  the  base  and  the  vertex  of  the 
angle  opposite  the  base.     Thus,  in  Fig.  28, 

.    =  AC  X  BD      BC  X  AF  _  AB  X  CE 

2       "  ~  2  2 

Here  AF  is  the  altitude  when  BC  is  the  base,  and  CE  is  the 

altitude  when  AE  is  the  base. 

In  a  right  triangle,  if  one  leg  is  taken  as  the  base,  the  other  leg 

is  the  altitude,  see  Fig.  22. 
Consequently,  the  area  of 
a  right  triangle  is  equal  to 
half  the  product  of  its  legs. 
If  two  triangles  have  the 
same  base  and  the  same 
altitude,  their  areas  are 
equal.  Thus,  in  Fig.  28, 
if  MN  is  parallel  to  AC, 

the  altitudes  of  the  three  triangles  ABC,  AMC,  and  ANC  are  equal; 

and  since  they  all  have  the  same  base  AC,  all  three  triangles  have 

the  same  area. 

Example.— If  the  base  of  a  triangle  is  2  ft.  9  in.  and  the  altitude  is  1  ft. 
4  in.,  what  is  the  area  of  the  triangle? 


§2  MENSURATION  OF  PLANE  FIGURES  57 

Solution. — The  lengths  must  be  expressed  in  the  same  single  unit,  ex- 
pressing them  in  feet,  2  ft.  9  in.  =  2.75  =  2 J  ft.;  1  ft.  4  in.  =  1J  ft. 
Then,  applying  the  formula,  A  =  (2.75  X  11)  +  2  =  1.8  \  Pq.  ft.  Arts. 

Or,   expressing  the  lengths  in  inches,   2  ft.  9  in.  =  33  in.;  1  ft.  4  in. 

=  16  in.     Then,  A  =  33  *  16  =  264  sq.  in.  =  ~|  =  If  =  1.81  sq.  ft.  Am. 

Always  remember  that  feet  multiplied  by  feet  give  square  feet 
and  inches  multiplied  by  inches  give  square  inches. 

69.  If  the  triangle  is  isosceles  (an  equilateral  triangle  is  also 
isosceles),  and  the  unequal  side  is  taken  as  the  base,  the  line  repre- 
senting the  altitude  bisects  the  base,  see  (5)  of  Art.  63.  Hence, 
referring  to  Fig.  19,  if  the  three  sides  a,  b,  and  c  are  known,  a 

and    b    being   equal,   the  altitude  CD    =    h    =    yl  b2  —  y^J 

=  h\/ib2  -  c2,  by  formula  (4)  or  (3),  Art.  66;  and  the  area  of  the 
triangle  is, 

A  =  \  X  c  X  i\/462  -  c2  =  |V462  -  c2. 

Example. — In  an  equilateral  triangle,  the  length  of  each  side  is  7%  in. ; 

what  is  the  area  of  the  triangle? 

7  25    /- ■ 

Solution.— In   this   case,    b  =  c;   hence,   A  =  -jv4  X  7.252  -  7.252 

=  ^V3  X  7.252  =  ^V3  =  22.760+  sq  in.     Ans. 
4  4 

For  purposes  of  reference,  let  a  =  one  of  the  equal  sides  of  an 
isosceles  triangle,  and  let  c  be  the  unequal  side;  then 

A  =  jV4a2-  c2  (1) 

For  an  equilateral  triangle,  let  c   =   one  of  the  sides;  then 
A  =  |V3  =  .43301c2  (2) 

70.  If  all  three  sides  of  any  triangle  are  known  and  the  altitude 
is  not  known  and  cannot  conveniently  be  measured,  let  p  =  a 
+  b  +  c;  that  is,  p  equals  the  sum  of  the  three  sides.     Then, 

A  =  \\pjp-  2a)(p-  2b)(p-2c) 
In  tins  formula,  a,  b,  and  c,  are  the  three  sides  of  the  triangle. 

Example. — Suppose  the  three  sides  of  a  triangle  are  6]  in.,  5f  in.,  and  8| 
in.;  what  is  the  area  of  the  triangle? 

Solution.— Here  p  =  6.5  +  5.75  +  8.125  =  20.375.  It  does  not  matter 
which  of  the  sides  are  designated  by  a,  by  b,  or  by  c;  hence,  taking  them  in 
the    order   given,   p  -  2a  =  20.375   -  2  X  6.5  =  7.375,    p  -  2b  =  20.375 


58 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


-  2  X  5.75  =  8.875,  and  p  -  2c  =  20.375  -  2  X  8.125  =  4.125.     Substi- 
tuting these  values  in  the  formula, 

A  =J\/20.375  X  7.375  X  8.875  X  4.125  =  18.542+.  Ans. 

When  the  triangle  is  equilateral,  a  =  b  =  c,  p  =  3a,  and  A 

=  I  \Sa{Sa  -2a)  (3a  -  2a)  (3a  -  2a)  =  \  \ka4  =  ^\S,  which 

is  the  same  as  formula  (2),  Art.  69,  when  a  is  substituted  for  c. 
Evidently  p  is  the  distance  around  the  triangle;  it  is  called  the 
perimeter,  peri  meaning  around  and  meter  meaning  measure. 
The  word  is  applied  to  the  distance  around  any  plane  figure ;  and 
when  the  plane  figure  is  a  polygon,  the  perimeter  of  the  polygon 
is  the  sum  of  the  lengths  of  its  sides. 

71.  Projections. — Hereafter,  unless  a  curve  or  broken  line  is 
specified,  the  word  line  will  be  understood  to  mean  a  right  line — 
usually,  a  right-line  segment. 

If  from  any  point  a  perpendicular  be  let  fall  upon  a  line,  the 
point  in  which  the  perpendicular  intersects  the  line  (the  foot  of 


the  perpendicular)  is  called  the  projection  of  the  point  upon  the 
line;  and  the  distance  from  the  point  to  its  projection  is  the  per- 
pendicular distance  between  the  point  and  the  line.  Thus,  in 
Fig.  29,  A',  B  ,  C,  and  D'  are  the  projections  of  A,  B,  C,  and  D 
upon  the  line  MN. 

The  perpendicular  distance  from  a  point  to  a  line  is  always  the 
shortest  distance  from  the  point  to  the  bne.  Thus,  if  from  B, 
any  line  BE  be  drawn,  BE  is  the  hypotenuse  of  a  right  triangle 
of  which  BB',  the  perpendicular  distance,  is  one  leg,  and  the 
hypotenuse  is  always  the  longest  side  of  a  right  triangle. 

If  two  points  of  any  line  are  projected  on  another  line  in  the 
same  plane,  the  line  segment  between  the  two  points  of  projection 


§2 


MENSURATION  OF  PLANE  FIGURES 


59 


is  called  the  projection  of  the  line  segment  between  the  two  points 
projected.  Thus,  in  Fig.  29,  A'B'  is  the  projection  of  AB  upon 
MN,  and  CD'  is  the  projection  of  CD  upon  MN.  Hence,  to 
project  any  line,  straight  or  curved,  upon  a  right  line  in  the  same 
plane,  project  its  two  end  points  upon  the  right  line,  and  the 
segment  included  between  the  two  points  of  projection  will  be  the 
projection  of  the  given  line.  In  Fig.  30,  A'B',  CD',  and  E'F' 
are  the  projections  of  AB,  CD,  and  EF  upon  the  line  MN. 


M- 


1' 


B! 


m     e' 


F' 


N 


Fig.  30. 


It  will  be  noted  that  the  projection  of  a  line  is  always  shorter 
than  the  given  line,  except  when  the  given  line  is  a  right  line 
parallel  to  the  line  of  projection;  in  which  case,  the  given  line 
and  the  projected  line  are  equal.  In  Fig.  30,  AB  is  parallel  to 
MN;  hence,  A'B'  =  AB. 

72.  Relation  between  Sides  of  any  Triangle. — In  Figs.  31 
and  32,  suppose  that  the  angle  A  is  acute.     Drop  a  perpendicular 


from  B  to  AC;  then  AD  is  the  projection  of  AB  upon  AC.  Rep- 
resent the  projection  AD  by  p.  In  Fig.  31,  DC  =  b  —  p,  and 
in  the  right  triangle  BDC,  BD2  =  a2  -  (b  -  p)2  =  a2  -  b2  + 
2bp  —  p2.  In  the  right  triangle,  BDA,  BD2  =  c2  —  p2;  therefore, 
BD2  =  a2  —  b2  +  2bp  —  p2  =   c2  —  p2,  or  transposing  terms, 

a2    =    52  _|_   C2   _   2bp        (1) 


60  ELEMENTARY  APPLIED  MATHEMATICS  §2 

In  Fig.  32,  DC  =  p  —  b,  and  in  the  right  triangle  BDC, 
BD2  =  a2  -  (p  -  b)-  =  a-  -  p2  +  2pb  -  b2.  In  the  right 
triangle  BDA,  BD2  =  c2  -  p2;  therefore,  BD2  =  a2  -  p2 
+  2bp  -  b2  =  c2  -  p2;  from  winch,  a2  =  b2  +  c2  -  2bp,  which  is 
the  same  as  (1). 

Stated  in  words,  the  square  of  the  side  opposite  an  acute  angle 
of  any  triangle  is  equal  to  the  sum  of  the  squares  of  the  other  two 

sides  minus  twice  the  product 
obtained  by  multiplying  one  of 
these  sides  by  the  projection  of 
the  other  side  upon  that  side. 
Thus,  in  Fig.  31,  the  projec- 
tion of  AC  upon  AB  is  AD' 
=  p',  and  a2  =  b2  +  c2  -  2cp'. 
This,  evidently,  is  the  same 
case  as  Fig.  32,  when  the  tri- 
angle is  turned  over  and  AB 
FlG  33  of  Fig.  31  is  made  the  base. 

In  Fig.  33,  DA  =  p  +  b,  p 
being  the  projection  DC  of  a  upon  AC.  Then,  BD2  =  a2  —  p2 
=  c2  —  (b  +  p)2  =  c2  —  b2  —  2bp  —p2;  from  which, 
c«-  =  a2  +  b2  +  2bp  (2) 
Stating  formula  (2)  in  words,  the  square  of  the  side  opposite  an 
obtuse  angle  of  any  triangle  is  equal  to  the  sum  of  the  squares  of  the 
other  two  sides  plus  twice  the  product  obtained  by  multiplying  one 
of  these  sides  by  the  projection  of  the  other  side  upon  hat  side.  Thus, 
in  Fig.  33,  the  projection  of  b  upon  BC  is  CD'  =  p',  and  c2  =  a2 
+  62  +  2a//. 

Example. — In  the  triangle  ABC,  in  which  the  side  AB  is  the  longest, 
AB  =  22  in.,  AC  =  13  in.,  and  the  projection  of  AC  upon  AB  measures 
9.73  in.,  what  is  the  length  of  the  side  CB? 

Solution. — The  sides  AB  and  AC  evidently  include  an  acute  angle, 
since  the  longest  side  is  always  opposite  the  largest  angle;  hence,  formula 
(1)  applies  to  the  case,  substituting  c  for  b,  so  that  the  formula  becomes 
a1  =  b*  +  cs  —  2cp;  whence, 

a  =  V&*  +  c2  -  2cp  =  V132  +222  -2  X22  X  9.73 

=  AB  =  15.029  in.     Ans. 

If  the  triangle  is  a  right  triangle,  the  projection  of  one  leg  upon 
the  other  is  a  point,  which  has  no  length  and  is  therefore  0.  In 
this  case,  c2  =  a2  +  b2  +  26  X  0  =  a2  +  b2,  which  is  the  same 
as  formula  (1)  in  Art.  66. 


§2 


MENSURATION  OF  PLANE  FIGURES 


61 


QUADRILATERALS 

73.  The  Parallelogram. — When  the  opposite  sides  of  a  quadri- 
lateral are  parallel,  the  quadrilateral  is  called  a  parallelogram; 
Fig.  34  shows  two  parallelograms  ABDC,  the  sides  AB  and  CD 
being  parallel  and  equal,  and  the  sides  AC  and  BD  being  also 
parallel  and  equal.        * 


When  the  interior  angles  are  not  all  equal  and  the  sides  are  not 
all  equal  the  parallelogram  is  called  a  rhomboid;  both  parallelo- 
grams in  Fig.  34  are  rhomboids. 

If  the  angles  are  not  all  equal,  but  the  sides  are  equal,  the  paral- 
lelogram is  called  a  rhombus;  see  Fig.  35,  in  which  AB  =  BD 
=  DC  =  CA. 

The  diagonal  of  a  parallelogram  is  a  line  drawn  through  the 
figure  from  the  vertex  of  one  acute  angle  to  the  vertex  of 
the  other  acute  angle,  the  line  CB  in  Figs.  34  and  35;  this  is  called 
the  long  diagonal.  A  line  AD,  drawn  from  the  vertex  of  one  ob- 
tuse angle  to  the  vertex  of  the  other  obtuse  angle,  Figs.  34  and 
35,  is  called  the  short  diagonal. 

The  perpendicular  distance  BE  between  two  parallel  sides,  in 
Figs.    34    and    35,    is    called    the 
altitude  of  the  parallelogram. 

74.  Some  Properties  of  Paral- 
lelograms.—(1)  The  diagonally 
opposite  angles  of  any  parallelo- 
gram are  equal;  thus,  in  Figs.  34 
and  35,  C  =  B  and  A  =  D.  This 
is  a  consequence  of  Art.  67,  since 
AB  is  parallel  to  CD  and  AC  is  parallel  to  BD. 

(2)  A  diagonal  divides  the  parallelogram  into  two  equal  tri- 
angles. Thus,  ACB  =  CDB  and  ACD  =  ABD.  This  isevident 
since  the  three  sides  of  one  triangle  are  equal  to  the  three  sides 
of  the  other,  the  diagonal  being  a  common  side. 

(3)  The  diagonals  of  a  parallelogram  bisect  each  other;  that 


Fig.  35. 


62 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


is,  P  being  the  point  of  intersection  of  the  diagonals,  PA  =  PD 
and  PB  =  PC. 

(4)  The  sum  of  the  interior  angles  of  any  parallelogram  is 
equal  to  4  right  angles  or  360°.  For,  referring  to  Figs.  34  and 
35,  angle  BDE  =  angle  ACD,  and  CDB  +  BDE  =  CDB  +  ACD 
=  C  +  D  =  2  right  angles  =  180°.  Since  A  =  D  and  B  =  C, 
i+5  +  D  +  C  =  2X  180°  =  360°.  Therefore,  if  one  angle 
of  a  parallelogram  is  known,  all  are  known.  Thus,  suppose  the 
angle  C  in  Fig.  35  is  55°;  then  angle  D  is  180°  -  55°  =  125°, 
A  =  125°,  and  B  =  55°. 

76.  The  side  on  which  the  parallelogram  is  supposed  to  stand 
is  called  the  base;  in  Figs.  34  and  35,  CD  is  the  base. 

The  area  of  any  parallelogram  is  equal  to  the  product  of  the 
base  and  altitude;  thus,  the  areas  of  the  parallelograms  in  Figs, 
34  and  35  is  equal  to  CD  X  BE.  Let  A  =  the  area,  I  =  the 
length  of  the  base,  and  h  =  the  altitude;  then, 

A  =  Ih. 

This  follows  at  once  from  the  rule  for  finding  the  area  of  a 
triangle.  Thus,  area  of  triangle  CDB,  Figs.  34  and  35,  equals 
Y2  X  BE  X  CD,  and  since  CAB  =  CDB,  the  area  of  the  paral- 
lelogram is  2  X  lA  CD  X  BE  =  CD  X  BE. 


76.  If  one  angle  of  a  parallelogram  is  a  right  angle,  all  the 
angles  are  right  angles,  in  accordance  with  Art.  74,  and  the  paral- 
lelogram is  then  called  a  rectangle;  see  (a),  Fig.  36.  Here  A 
=  B  =  C  =  D  =  &  right  angle  =  90°,  and  BC  =  AD  and  AB  = 
DC. 

If  the  four  sides  of  a  rectangle  are  equal,  it  is  called  a  square. 
Thus,  (b),  Fig.  36,  is  a  square,  because  AB  =  BC  =  CD  =  DA 
andA  =  £  =  C  =  Z)  =  a  right  angle. 


§2 


MENSURATION  OF  PLANE  FIGURES 


63 


Referring  to  (a),  Fig.  36,  it  will  be  noted  that  if  AD  is  the  base, 
CD  =  AB  is  the  altitude,  and  if  AB  is  the  base,  AD  =  BC  is 
the  altitude.  Hence,  if  one  side  of  a  rectangle  be  called  the  length, 
and  one  of  the  sides  perpendicular  to  it  be  called  the  breadth  or 
depth  (according  to  whether  the  plane  of  the  rectangle  is  sup- 
posed to  be  horizontal  or  vertical),  the  area  of  the  rectangle  is 
equal  to  the  product  of  the  length  and  breadth  or  depth;  that  is, 

A  =  lb  =  Id. 
in  which  A  =  the  area,  I  =  AD,  and  b  =  d  =  AB  =  DC. 

When  the  rectangle  is  a  square,  I  =  b  =  d,  and  A  =  I  X  I  =  Z2. 

77.  Referring  to  Fig.  37,  A  BCD  is  a  rhombus.  Draw  the  two 
diagonals,  and  represent  the  long  diagonal 
by  d  and  the  short  diagonal  by  d'.  The 
diagonals  intersect  in  P,  which  is  the  middle 
point  of  BD  and  AC  (see  Art.  74),  and  the 
four  angles  about  P  are  right  angles.  That 
BPA  =  BPC  is  a  right  triangle  is  evident 
from  (5),  Art.  63.  Here  ABC  is  an  isosceles 
triangle  and  since  P  is  the  middle  point  of 
the  base,  the  line  from  P  to  B  is  perpen- 
dicular to  the  base  (see  Fig.  19).  Con- 
sidering the  triangle  DAB  (=  DCB),  AP 
is  the  altitude  and  DB  is  the  base.  But 
d' 


Fig.  37. 


AP  =  2  and  DB 


d;  hence,  area  of  DAB  =  %  X  d  X  ~z}  and 


area  of  rhombus  A  BCD  is 


2 


A  =\XdX^X2  =  \dd'- 


(1) 


sides ; 


That   is,   the  area  of  a  rhombus  is  equal  to 
one-half  the  product  of  its  diagonals. 

Referring  to  Fig.  38,  A  BCD  is  a  square — 
a  rhombus  whose  angles  are  right  angles. 
The  diagonals  AC  and  BD  are  equal  and 
formula  (1)  becomes 

A  =  \dd  =  \d2.  (2) 

If  it  is  desired  to  find  the  length  of  the 
diagonal   of   any    square,  let  I  =  one  of  the 
then,  d2  =  I2  +  I2  =  2l2,  and 

d  =  VW  =  lV2  =  1.41422.  (3) 


Fig.  38. 


64  ELEMENTARY  APPLIED  MATHEMATICS 


§2 


If  the  diagonal  of  a  square  is  given  and  it  is  desired  to  find  the 
length  of  the  sides,  I2  +  I2  =  d2  or  2/2  =  d2,  and  I2  =  -gi 

from  which  I  =  d2  a/2  =  .707107d         (4) 

Example  1. — Fig.  37  is  a  drawing  of  a  flat,  rectangular  plate  having  a 
hole  in  it  shaped  like  a  rhombus.  From  the  dimensions  given,  find  the 
area  of  the  plate. 

Solution. — The  area  of  the  plate  is  evidently  equal  to  the  area  of  the 
rectangle  minus  the  area  of  the  rhombus. 

Area  of  rectangle  =  18.5  X  28.625  =  529.5625  sq.  in. 

Area  of  rhombus  =  M  X  10.375  X  12.75  =    66.1406  sq.  in. 
area  of  plate  =  463.4219  sq.  in. 

Therefore,  area  of  plate  is  463.92  sq.  in.     Arts. 

Example  2. — What  is  the  diagonal  of  a  square,  one  side  of  which  measures 

9^6  in.? 
Solution. — Substituting  in  formula  (.3), 

d  =  1.4142  X  9^6  =  13.170-  in.     Ans. 
Example  3. — If  the  diagonal  of  a  square  is  11.82  in.,  what  is  the  length  of 
one  of  the  sides? 

Solution. — Applying  formula  (4), 

I  =  .707107  X  11.82  =  8.358  in.     Ans. 

78.  Other  Quadrilaterals. — If  two  sides  of  a  quadrilateral  are 
parallel  and  the  other  two  sides  are  not  parallel,  the  quadrilateral 
is  called  a  trapezoid.     In  Fig.  39,  (a)  and  (6)  are  trapezoids,  the 


(b) 


D 


side  BC  being  parallel  to  AD.     The  trapezoid  in  (6)  is  peculiar 
from  the  fact  that  it  has  two  right  angles,  situated  at  A  and  B. 

If  no  two  sides  of  a  quadrilateral  are  parallel,  it  is  then  called 
a  trapezium ;  see  Fig.  40. 

To  rind  the  area  of  a  trapezoid,  divide  it  into  two  triangles  by 
diawing  a  diagonal  AC;  then,  if  BC  be  taken  as  the  base  of  one 
triangle  and  AD  as  the  base  of  the  other,  the  altitude  of  these 
triangles  is  AD  =  CE  in  (a),  Fig.  39,  and  AB  =  the  perpendicular 
distance  between  the  parallel  sides  in  {b),  Fig.  39.     Denoting  the 


§2  MENSURATION  OF  PLANE  FIGURES  65 

altitude  by  h,  the  side  BC  by  a,  and  the  side  AD  by  b,  the  area 
of  one  triangle  is  }iah;  of  the  other,  }ibh;  and  the  area  of  the 
trapezoid  is  3^  ah  +  xAbh  =  }ih(a  +  6);  or, 

a  =  !l+ 6  x  h  =  fe±*& 

This  is  a  very  important  formula;  it  is  frequently  used,  and 
should  be  carefully  committed  to  memory.  Stated  in  words,  the 
formula  becomes  the  following  rule:  The  area  of  a  trapezoid  is 


Fig.  40. 

equal  to  half  the  sum  of  the  parallel  sides  multiplied  by  the  perpen- 
dicular distance  between  them. 

Example. — What  is  the  area  of  a  trapezoid,  the  lengths  of  the  parallel 

sides  being  63^  in.  and  7}i  in.  and  the  altitude  being  9%  in.? 

Solution. — The  sum  of  the  parallel  sides  is  6.5  +  7.25  =  13.75;  hence, 

.,              .    13.75  X  9.375       _.  .__  ,  .         . 

the  area  is ^ =  64.453+  sq.  in.     Ans. 

It  is  better  not  to  divide  by  2  until  after  the  multiplication  has 
been  performed,  unless  one  of  the  factors  is  an  even  number.  In 
the  example  just  given,  both  factors  were  odd  numbers. 

79.  To  find  the  area  of  a  trapezium,  draw  a  diagonal,  as  BD, 
Fig.  40,  thus  dividing  the  trapezium  into  two  triangles.  Using 
the  diagonal  as  a  base,  draw  the  perpendiculars  AD  and  CE 
from  the  vertexes  opposite  the  base;  then,  area  of  trapezium  is 
equal  to  V2BD  X  EC  +  V2BD  X  AD  =  Y2BD  (AD  +  EC). 

If  through  one  of  the  vertexes,  say  C,  a  line  be  drawn  parallel 
to  the  base  (diagonal)  BD  and  a  perpendicular  be  drawn  to  this 
line  from  the  other  vertex,  the  length  of  this  perpendicular  is 


66  ELEMENTARY  APPLIED  MATHEMATICS  §2 

equal  to  AD   +   EC.     Representing  this  perpendicular  by  h, 
the  base  (diagonal)  by  b,  and  the  area  by  A, 

A  =  y2bh. 

80.  The  rules  and  formulas  just  given  for  finding  the  areas 
of  quadrilaterals  apply  only  to  what  are  termed  convex  polygons. 
All  polygons  that  have  been  illustrated  up  to  this  point  are 
convex  polygons;  and  in  any  convex  polygon,  if  any  one  of  their 
sides  be  produced  at  either  end  of  the  side,  the  produced  part  will 
lie  outside  of  the  bounding  line  of  the  polygon.  In  Fig.  41,  (a) 
is  a  quadrilateral  and  (6)  is  a  pentagon.     If  the  sides  AB  or 


CD  be  produced  from  the  end  B,  they  will  enter  the  space  included 
by  the  bounding  lines  of  the  polygons.  Angles  like  ABC  are 
called  re-entrant  angles,  and  polygons  having  one  or  more  re- 
entrant angles  are  called  concave  or  re-entrant  polygons. 

To  find  the  area  of  a  re-entrant  polygon,  divide  it  into  triangles, 
as  indicated  by  the  dotted  lines,  find  the  area  of  each  triangle, 
and  their  sum  will  be  the  area  of  the  polygon. 

Unless  otherwise  stated,  all  polygons  are  supposed  to  be  convex 
polygons. 


REGULAR  POLYGONS 

81.  Except  in  the  case  of  irregular  figures  bounded  by  right 
lines,  most  of  the  polygons  that  occur  in  practice  are  regular 
polygons. 

If,  in  any  regular  polygon  having  an  even  number  of  sides,  a  line 
(diagonal)  be  drawn  from  any  vertex*to  the  vertex  opposite  that 
is  farthest  away,  the  line  will  pass  through  what  is  called  the 
geometrical  center  of  the  polygon;  and  if  two  such  lines  be  drawn 


§2 


MENSURATION  OF  PLANE  FIGURES 


67 


from  different  vertexes,  they  will  intersect  in  the  geometrical 
center.  Thus,  in  Fig.  42,  which  represents  a  hexagon  (a  polygon 
with  an  even  number  of  sides),  AD,  BE,  and  CF  all  intersect  in  0, 
which  is  the  geometrical  center,  or,  more  simply,  the  center, 
of  the  hexagon. 

If  a  regular  polygon  have  an  odd  number  of  sides,  as  the  pen- 
tagon, Fig.  43,  and  a  perpendicular  be  drawn  from  any  vertex 
to  the  side  opposite,  it  will  pass  through  the  geometrical  center 
of  the  polygon;  and  any  two  such  perpendiculars  will  intersect 
in  the  center.  Thus,  AP,  BQ,  CR,  DS,  and  ET,  which  are  per- 
pendicular respectively  to  CD  DE,  EA,  AB,  and  BC,  all  inter- 
sect in  0,  the  center  of  the  pentagon. 


The  perpendicular  from  the  center  to  one  of  the  sides,  as  OP 
in  Figs.  42  and  43,  bisects  the  side;  that  is,  PC  =  PD.  This 
perpendicular  is  called  the  apothem. 

The  lines  AD,  BE,  etc.,  Fig.  42,  and  AP,  BQ,  etc.,  Fig.  43, 
divide  the  polygons  into  as  many  equal  triangles  as  the  poly- 
gons have  sides  and  the  sum  of  the  areas  of  these  triangles 
equals  the  areas  of  the  polygons.  The  area  of  one  triangle  is, 
letting  I  =  length  of  one  side  and  a  =  the  apothem,  }4aL 
If  n  =  the  number  of  sides  of  the  polygon,  the  area,  A,  of  the 
polygon  is  %  anl.  But  nl  —  the  perimeter  of  the  polygon  =  p; 
hence, 

A  =  Y2Va.  (1) 

Stated  in  words,  the  area  of  any  regular  polygon  is  equal  to  half  its 
perimeter  multiplied  by  its  apothem. 


lis 


elementary  applied  mathematics 


§2 


TABLE  OF  REGULAR  POLYGONS 


Number  of 

Apothcm 

Area 

Number  of 

Apotbem 

Area 

sides 

a 

k 

sides 

a 

k 

3 

0.28868 

0.43301 

15 

2.3523 

17.642 

4 

0.5 

1. 

16 

2.5137 

20.109 

5 

0.68819 

1 . 7205 

20 

3.1569 

31 . 569 

6 

0.86603 

2.5981 

24 

3.7979 

45.575 

7 

1.0383 

3.6339 

25 

3.9579 

49.474 

8 

1.2071 

4.8284 

30 

4.7572 

71 . 358 

9 

1.3737 

6.1818 

32 

5.0766 

81.225 

10 

1 . 5388 

7.6942 

40 

6.3531 

127.06 

11 

1.7028 

9.3656 

48 

7.6285 

183.08 

12 

1.8660 

11.196 

64 

10.178 

325.69 

The  area  may  also  be  found  by  means  of  the  above  table  when 
the  number  of  sides  in  the  polygon  is  given  in  the  table.  To  use 
the  table,  let  A;  =  the  number  in  the  column  headed  area  that 
coresponds  to  the  given  number  of  sides;  let  I  =  the  length  of 
the  given  side;  then, 

A  =  kl\  (2) 

For  example  suppose  that  the  length  of  a  side  of  a  regular 
octagon  is  3)^  in.,  and  it  is  desired  to  find  the  area  of  the  octagon. 
Referring  to  the  table,  when  the  number  of  sides  is  8,  k  =  4.8284 ; 
hence,  the  area  is  A  =  4.8284  X  3.252  =  51  sq.  in.     Ans. 

Example. — One  side  of  a  hexagonal  bar  of  iron  measures  1H  in.;  what 
is  the  area  of  a  cross  section  of  the  bar? 

Solution. — Referring  to  the  table,  when  the  number  of  sides  is  6,  k 
-  2.5981;  hence,  area  =  A  =  2.5981  X  (1H)2  =  7.3985-,  say  7.398  sq. 
in.     Ans. 

The  apothem  can  be  used  to  lay  out  the  polygon ;  the  manner 
of  doing  this  will  be  described  later.  The  apothem  as  given  in 
the  table  is  for  a  side  equal  to  1 ;  hence  to  find  the  actual  length 
of  the  apothem  when  the  length  of  a  side  of  the  polygon  is  given 
let  I  =  length  of  side;  then  actual  length  of  apothem  =la.  In 
the  last  example,  the  actual  length  of  the  apothem  is  1-U 
X  .86603  =  1.4614  in. 


THE  CIRCLE 

82.  Definition. — The  circle  is  a  curve  every  point  of  which  is 
equally  distant  from  a  point  within  it  called  the  center.     The 


§2 


MENSURATION  OF  PLANE  FIGURES 


69 


curve  shown  in  Fig.  44  is  a  circle,  0  being  the  center.  A  circle 
may  be  described  (drawn)  in  various  ways.  Thus,  with  a  pin, 
punch  two  holes  in  a  strip  of  heavy 
paper;  put  the  pin  through  one 
hole  and  the  point  of  a  sharp  pencil 
through  the  other  hole;  then,  keep- 
ing the  pin  stationary,  revolve  the 
pencil  about  the  pin,  keeping  the 
strip  of  paper  stretched  tight,  and 
the  pencil  will  describe  a  circle. 
This  method  is  shown  in  Fig.  45. 
A  better  way  is  to  use  an  instru- 
ment employed  by  draftsmen, 
called  compasses.  This  instru- 
ment consists  of  two  legs  united  at 
one  end  by  a  joint,  which  permits  them  to  open  and  close  to  any 
desired  distance  apart.     One  leg  has  a  needle  point  at  one  end, 


Fig.  45. 


and  the  other  leg  carries  a  pencil  point  or  pen.     By  placing  the 
needle  point  of  the  compasses  at  the  center,  the  leg  carrying  the 


70  ELEMENTARY  APPLIED  MATHEMATICS  §2 

pencil  point  or  pen  may  be  revolved  about  the  needle  point, 
thus  describing  the  circle  as  shown  in  Tig.  46. 

83.  Referring  to  Fig.  44,  any  part  of  a  circle,  as  AD,  ABC,  etc. 
is  called  an  arc  of  a  circle,  a  circular  arc,  or  simply,  an  arc; 
it  is  so  called  from  its  shape,  the  word  arc  meaning  bow.  A 
right  line  joining  the  extremities  of  an  arc  is  called  the  chord 
of  the  arc  or,  simply,  the  chord.  Thus,  AC  is  the  chord  of  the 
arc  ABC,  and  DE  is  the  chord  of  the  arc  DABCE.  When 
the  chord  passes  through  the  center  of  the  circle,  it  divides  the 
circle  into  two  equal  parts,  each  of  which  is  called  a  semicircle, 
meaning  half-circle,  and  the  chord  is  then  called  a  diameter  of 
the  circle  or,  simply,  a  diameter.  In  Fig.  44,  DE  is  a  diameter 
of  the  circle,  because  it  passes  through  the  center.  For  the  same 
reason,  BF  is  also  a  diameter.  The  arc  DBE  is  equal  to  the  arc 
DFE,  and  both  are  semicircles.  The  arc  FDB  is  equal  to  the 
arc  FEB,  and  both  of  these  arcs  are  semicircles. 

A  right  line  drawn  from  the  center  to  the  curve  is  called  a 
radius  of  the  circle;  thus,  OD,  Fig.  44,  is  a  radius,  and  so  is  OA, 
OB,  etc.  The  plural  of  radius  is  radii;  hence,  OD,  OA,  OB,  OC, 
OE,  and  OF  are  radii,  of  the  circle  DABCEF.  All  radii  of  any 
circle  are  equal,  by  definition  of  the  circle,  since  they  equal  the 
distance  from  the  center  to  the  curve.  A  radius  is  also  equal  to 
one-half  the  diameter,  since  the  diameter  DE  =  OE  +  OD,  and 
OE  =  OD  =  the  radius.  Consequently,  the  diameter  equals 
twice  the  radius. 

The  perimeter  of  a  circle  is  commonly  called  the  circumference ; 
in  geometry,  it  is  called  the  periphery.  The  word  periphery 
is  applied  to  plane  figures  having  curved  outlines,  while  the  word 
perimeter  is  applied  to  plane  figures  bounded  by  right  lines. 

84.  The  word  circle  is  also  applied  to  the  area  contained  within 
the  circumference,  hence,  by  area  of  a  circle,  the  area  included  by 
the  circumference  is  always  meant.  The  area  included  by  an 
arc  and  two  radii  drawn  to  the  extremities  of  the  arc  is  called  a 
sector;  in  Fig.  44,  the  area  OABC  is  a  sector  of  the  circle  DBEF. 
The  area  included  between  an  arc  and  its  chord  is  called  a 
segment;  the  area  ABC  A  is  a  segment  of  the  circle  DBEF. 

85.  If  a  line  be  drawn  from  a  point  without  a  circle  and  is 
terminated  by  the  circumference  after  passing  through  the  cir- 
cle, such  a  line  is  called  a  secant.  In  Fig.  47,  PA  and  PB  are  se- 
cants.    Evidently,  a  secant  intersects  the  circumference  in  two 


§2 


MENSURATION  OF  PLANE  FIGURES 


71 


Fig.  47. 


points;  thus,  PA  intersects  the  circumference  in  A  and  D,  and 
PB  intersects  it  in  B  and  E.  If,  however,  the  secant  just  touches 
the  circle,  intersecting  it  in  only  one  point,  it  is  called  a  tangent; 
thus,  PC  is  a  tangent,  because  it  intersects  the  circle  in  only  one 
point,  the  point  C,  which  is  called  the 
point  of  tangency.  PG  is  also  a  tangent, 
and  F  is  the  point  of  tangency. 

86.  Some  Properties  of  Circles. — (1) 
If  a  diameter  be  drawn  perpendicular  to 
any  chord,  it  bisects  the  chord  and  also 
the  arc.  In  Fig.  44,  if  BF  is  perpen- 
dicular to  AC,  AG  =  GC,  and  arc  AB 
=  arc  BC. 

(2)  Any  angle  whose  vertex  is  the  center 
of  the  circle  is  measured  by  the  arc  it  inter- 
cepts; the  word  intercept  here  means  the 
part  of  the  circumference  cut  off  by  and  included  between  the 
radii  forming  the  sides  of  the  angle.  In  Fig.  44,  the  angle  COE  is 
measured  by  the  arc  CE  that  is  intercepted  by  the  radii  CO  and 
CE.  The  angle  AOC  is  measured  by  the  intercepted  arc  ABC; 
the  angle  DO  A ,  by  the  intercepted  arc  DA ;  etc  It  is  here  under- 
stood that  the  circumference  is  supposed  to  be  divided  into  de- 
grees, minutes,  and  seconds,  as  described  in  Art.  54. 

Angles  whose  vertexes  are  situated  at  the  center  of  the  circle 
are  called  central  angles  or  angles  at  the  center. 

(3)  Since  central  angles  are  measured  by  the  arcs  they  inter- 
cept, it  is  evident  that  a  diameter  perpendicular  to  the  chord  of 
the  arc  bisects  the  central  angle  that  is  measured  by  that  arc. 
Thus,  by  (1),  if  the  diameter  BF  is  perpendicular  to  AC,  arc 
AB  =  arc  BC,  and  AOB  =  BOC,  since  both  angles  are  measured 

by  equal  arcs. 

(4)  If  a  right  line  be  drawn  perpendicular  to  any  chord  at  its 
middle  point,  it  will  pass  through  the  center  of  the  circle  having  the 
same  arc  that  is  subtended  by  the  chord.  In  Fig.  44,  the  arc  ABC 
is  subtended  by  the  chord  AC.  If  G  is  the  middle  point  of  the 
chord  and  BGF  is  perpendicular  to  AC,  then  BGF  must  pass 
through  the  center  0  of  the  circle  DBEF  of  which  the  arc  ABC 

is  a  part. 

(5)  Two  circles  are  equal  when  the  radius  or  diameter  of  one  is 
equal  to  the  radius  or  diameter  of  the  other;  two  sectors  are  equal 
when  the  radius  and  chord  of  one  are  equal  to  the  radius  arid  chord 


72 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


of  the  other;  and  two  segments  or  two  arcs  are  equal  when  the  radius 
and  chord  of  one  are  equal  to  the  radius  and  chord  of  the  other. 

(6)  //  the  vertex  of  an  angle  lies  on  the  circumference,  the  angle 
is  measured  by  one-half  the  intercepted  arc.  In  Fig.  48,  the  vertex 
of  BAC  lies  on  the  circumference;  it  is  therefore  measured  by  one- 
half  the  arc  BC.     Angles  whose  vertexes  lie  on  the  circumference 


Fig.  48. 


Fig.  49. 


are  called  inscribed  angles;  hence,  an  inscribed  angle  is  one-half 
as  large  as  the  central  angle  having  the  same  arc.  Thus,  BAC 
=  y2  BOC,  and  BOC  =  2  BAC. 

(7)  If  the  inscribed  angle  intercepts  a  semicircle,  the  angle  is  a 
right  angle,  since  a  semicircle  contains  360°  ■*■  2  =  180°,  and  one- 
half  of  180°  is  90°,  a  right  angle.  Thus,  in  Fig.  49,  if  AC  is  a 
diameter,  ABC,  ADC,  and  AEC  are  all  right  angles.     Hence, 


Fig.  50. 

any  angle  inscribed  in  a  semicircle  is  a  right  angle.  This  fact  is 
made  use  of  by  mechanics  to  test  the  roundness  of  a  semicircular 
hole,  as  shown  in  Fig.  50.  Here  a  square  is  laid  across  the  edges, 
and  is  then  rotated  back  and  forth.  If  the  sides  of  the  square 
just  touch  the  edges  and  the  point  of  the  square  just  touches  the 
bottom,  the  surface  touched  is  semicircular,  since,  as  shown  in 


§2 


MENSURATION  OF  PLANE  FIGURES 


73 


the  figure,  ABC  is  a  semicircle  and  the  angle  ABC  inscribed  in  it 
is  a  right  angle. 

(8)  If  two  chords  intersect  in  a  point  within  a  circle,  as  DE  and 
FG,  Fig.  48,  which  intersect  in  H ,  the  angle  GHE,  which  equals 
DHF,  is  measured  by  one-half  the  sum  of  the  arcs  intercepted  by 
these  equal  angles;  that  is,  GHE  (or  DHF)  is  measured  by  one- 
half  of  arc  GE  +  arc  DF. 

(9)  If  two  secants  are  drawn  from  the  same  point,  the  angle 
between  the  secants  is  measured  by  one-half  the  difference  of 
the  arcs  they  intercept.  In  Fig.  47,  APB  is  measured  by  one- 
half  of  arc  AB  —  arc  DE. 

(10)  The  angle  between  a  secant  and  a  tangent  drawn  from  the 
same  point,  also  the  angle  between  two  tangents  drawn  from  the 
same  point,  is  measured  by  one-half  the  difference  of  the  inter- 
cepted arcs.  In  Fig.  47,  the  angle  APF  is  measured  by  one-half 
of  arc  AF  —  arc  FD,  and 
angle  FPC  is  measured  by 
one-half  arc  of  FBC  —  arc 
FEC. 

(11)  The  radius  drawn  to 
the  point  of  tangency  is  per- 
pendicular to  the  tangent. 
In  Fig.  47,  if  C  and  F  are 
points  of  tangency,  OC  is 
perpendicular  to  PC  and 
OF  is  perpendicular  to  PF. 

(12)  If  two  circles  in- 
tersect, the  line  passing 
through  their  centers  is 
perpendicular  to  their  com- 
mon chord  and  bisects  the 
chord.  In  Fig.  51,  two 
circles  having  the  centers 
O  and  0'  intersect;  AB 
is  their  common  chord,  drawn  through  the  points  of  intersec- 
tion; then,  the  right  line  passing  through  the  centers  0  and  0' 
is  perpendicular  to  A B  and  bisects  AB.  The  circles  whose 
centers  are  O  and  0"  also  intersect,  their  common  chord 
being  CD.  Then,  the  right  line  passing  through  0  and  0"  is 
perpendicular  to  CD  and  bisects  CD.  In  the  first  case,  the  center 
of  the  second  circle  is  outside  the  circumference  of  the  first ;  but, 


74 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


in  the  second  case,  the  center  of  the  second  circle  is  within  the 
circumference  of  the  first. 

(13)  From  a  given  point  without  a  circle,  two  tangents  may  be 
drawn  to  the  circle,  as  PF  and  PC  in  Fig.  47.  If  from  the  given 
point,  a  secant  be  drawn  passing  through  the  center,  it  bisects  the 
angle  formed  by  the  two  tangents  and  is  perpendicular  to  the 
chord  joining  the  points  of  tangency.  In  Fig.  47,  if  F  and  C 
are  the  points  of  tangency  of  the  tangents  drawn  from  P,  and 
PB  is  a  secant  passing  through  the  center  0,  then  OPF  =  OPC  = 
1  iFPC,  and  PB  bisects  the  chord  FC.  PB  is  also  perpendicular  to 
FC.     It  is  evident,  also,  that  PB  bisects  the  central  angle  FOC. 

87.  Three  Important  Principles. — (!)  If  any  two  chords  of  a 
circle  intersect,  the  product  of  the  segments  of  one  line  is  equal  to 
the  product  of  the  segments  of  the  other  line,  the  segments  being 
determined  by  the  point  of  intersection.  In  Fig.  52,  the  chords 
AB  and  CD  intersect  in  the  point  M;  then,  AM  X  MB  =  CM 

X  MD.  The  chords  AB 
and  JK  intersect  in  the 
point  X;  then,  AN  X  NB 
=  JX  X  NK. 

If  EF  is  a  diameter  per- 
pendicular to  the  chord 
GH,  IG  =  IH,  according  to 
(1),  Art.  86,  and  IE  X  IF 
=  IG  X  IH  =  IG2  =  IH2; 
that  is,  a  diameter  per- 
pendicular to  a  chord  is 
divided  by  the  chord  into 
two  segments  whose  pro- 
duct is  equal  to  the  square 
of  half  the  chord.  The 
segment  included  between  the  chord  and  the  arc  is  called  the 
height  of  the  arc  or  height  of  the  segment;  hence  if  the  chord 
and  height  of  the  arc  are  known,  the  diameter  or  radius  can  be 
found.     For,  let  c  =  the  chord  GH,  h  =  the  height  IE,  and  d  = 

the  diameter;  then  IF  =  d  -  h,   IG  =  ~  and  h  (d  -  h)  =  (~)  *; 

c2 
from  which,  hd  —  h2  =  —. -,  or 

4 

d  = 


tUL,. 

i 

\  TT 

I                                 \ 

o 

cl"^^-— JT^\ 

F 

Fig.  52. 


c2  +  Ah2 
4h 


(1) 


§2 


MENSURATION  OF  PLANE  FIGURES 


75 


Representing  the  radius  by  r,  r 
c2  +  4h2 


r  = 


8h 


,,  and 

(2) 


If  r  and  c  are  known  and  it  is  desired  to  find  h,  it  may  be 
found  from  formula  (2),  its  value  being 

h  =  r  ±  |  \/4r2  -  c2 
When  A  is  less  than  r,  the  minus  sign  is  used,  and 

h  =  r  -  iV4r2  -  c2  (3) 

When  /i  is  greater  than  r,  the  plus  sign  is  used,  and 

h  =  r  4-  ^V4r2  -  c2  (4) 

If  r  and  A  are  given  and  c  is  desired, 

c  =  2V(2r  -  A)A  (5) 

(2)  //  //-o/?i  a  pm'n<  without  a  circle  two  secants  are  drawn,  the 
product  of  the  whole  secant  and  the  external  segment  of  one  line  is 
equal  to  the  product  of  the  whole 
secant  and  the  external  segment 
of  the  other  line.  In  Fig.  53, 
PA  and  PB  are  secants  drawn 
from  the  point  P;  EP  and  FP 
are  the  external  segments; 
then,  PA  X  PE  =  PB  X  PF. 

(3)  If  from  a  point  without 
a  circle  a  secant  and  a  tangent 
are  drawn,  the  product  of  the 
whole  secant  and  its  external 
segment  is  equal  to  the  square 
of  the  tangent.  In  Fig.  53,  let 
PA  be  any  secant  and  PC  a 
tangent,  both  drawn  from  P; 
then,  PA  X  PE  =  PC2.  In 
order  to  measure  PC,  it  is 
necessary  to  know  the  point 
of  tangency  C;  and  this  can 
be  found  by  drawing  from 
the  center  0  a  perpendicular  to  PC,  the  point  where  it  intersects 
PC  being  the  point  of  tangency. 

Example.— The  chord  of  an  arc  has  a  length  of  14%  in.;  the  height  of  the 
arc  is  3K  in.;  what  is  the  radius? 


Fig.  53. 


76  KLEMENTARY  APPLIED  MATHEMATICS 


§2 


Solution.— Applying  formula  (2),  c  =  14.875,  h  =  3.25,  and 
14.875' +4  X3.25'=  .        Ans 

8  X  3.25 
88.  It  is  sometimes  desirable  to  know  the  chord  and  height 
of  half  the  arc  when  the  chord  and  height  of  the  whole  arc  are 
given.     In  such  a  case,  formulas  may  be  found  as    follows: 

Referring  to  Fig.  54,  let  C 
=  AB  and  H  =  CD,  the  chord 
and  height  of  the  arc  ACB; 
let  c  =  AC  and  h  =  EF,  the 
chord  and  height  of  the  arc 
AC  =  half  the  arc  ACB;  and 
let  r  =  the  radius  of  the  arc 
ACB.  In  the  right  triangle 
ADC,  AC2  =  AD2  +  CD2,  or 


(D 


+  H2  = 


C2  +  4H2 


Multiplying  and  dividing  this 
fraction  by  2H,  which  of 
course  does  not  alter  its  value, 
2H(C2  +  AH2) 


the   result    is 


C2  +  4H2 
8H      ' 


8H 
Therefore, 


=  2rH,  since  by  formula  (2),  Art.  87,  r  = 

c2  =  2rH,  and 

c  =  \/2lH  U) 

To  find  h,  EF  XFH  =  AF2,  or  h(2r  -  h)  =  (|) 
FII  =  2r-  h.     But  c2  =  2rH;  hence,  h(2r  -h)=^=~,  and 


since 


2rh  -h2  = 


rH 


Dividing  both  members  of  this  equation  by  —  1, 

to  change  the  sign  of  h2, 

rH 
2  ' 

Solving  this  equation  by  the  regular  rule  for  quadratics  (Art. 
31) 


h2  -  2rh  =  - 


2r 


h  = 


±v 


(2r)2  +  4  XIX- 


2  X  1 


-'±>H'-D- 


rH 

—  =  r  ±  \r 


(2) 


i-H 
~2 


§2  MENSURATION  OF  PLANE  FIGURES  77 

To  apply  either  of  these  formulas,  it  is  first  necessary  to  calcu- 
late the  radius.  ' 

Example. — Referring  to  the  example  in  Art.  87,  find  the  chord  and  height 
of  half  the  arc. 

Solution.— The  chord  AB  =  14.875  in.,  the  height  CD  =  3.25,  from 
which  the  radius  was  found  to  be  10.135  in.  By  formula  (1),  the  chord  of 
half  the  arc  =  c  =  y/2  X  10.135  X  3.25  =8.1165-,  say  8.116  in.     Ans. 

By    formula     (2),     h  =  10.135  ±  JlO.1352  -  10135  X  325   =  10.135 

+  9.287  =  19.422  in.;  or,  10.135  -  9.287  =  .848  in.     The  smaller  value  is 
evidently  the  one  required  in  this  case;  hence,  h  =  .848  in.     Ans. 

It  may  be  remarked  that  the  smaller  of  the  two  values  just  ob- 
tained is  the  length  of  EF,  while  the  larger  value  is  the  length 
FH\  the  sum  of  these  two  lengths  is  19.422  +  .848  =  20.270 
=  EH  =  2r  =  2  X  10.135  =  20.270. 

89.  Circumference  and  Area  of  the  Circle. — The  circumfer- 
ence of  a  circle  is  equal  to  the  diameter  multiplied  by  a  number 
that  is  universally  represented  by  the  Greek  letter  x  (pronounced 
pi  or  pe).  This  number  has  been  calculated  to  707  decimal 
places,  and  it  has  been  proven  that  it  cannot  be  expressed  by 
a  finite  number  of  figures;  its  value  to  9  significant  figures  is 
3.14159265  +  ,  and  is  usually  expressed  as  3.1416;  for  rough 
calculations,  3t  =  ~t~  is  commonly  used  for  t.  Letting  c  =  the 
circumference,  d  =  the  diameter,  and  A  =  the  area, 

c  =  wd  =  3.1416  d  (1) 

Since    the    diameter    equals    twice    the    radius, 

c  =  2irr  (2) 

The  area  of  a  circle  is  equal  to  t  times  the  square  of  the  radius, 
or 

A  =  Trr2  (3) 

Since  r  =  |,  A  =  t(|)2=  lird2  =  .7854d2  (4) 

These  four  formulas  are  extremely  important;  they  should  be 
carefully  committed  to  memory. 
From    formula    (1), 


rf  = ;  =  anile  -  -31831c       (5) 


From  formula  (2), 


r  =  ^  =  .159155c  (6) 

ii7r 


78  ELEMENTARY  APPLIED  MATHEMATICS  §2 

From  formula  (3), 

=  ^?  =  .56419  VA  (7) 


r 
From  formula  (5), 


4A 


=  1.1284  VA  (8) 


Formulas  (5)  to  (8)  may  be  used  to  calculate  the  radius  or 
diameter  when  the  circumference  or  area  is  known;  as  a  rule, 
however,  these  formulas  are  not  used,  formulas  (1)  to  (4)  being 
preferred. 

Example  1. — A  pulley  has  a  diameter  of  32  in.  and  makes  175  revolutions 
per  minute;  how  fust  does  a  point  on  the  rim  travel  in  feet  per  minute? 

Solution. — When  the  pulley  has  turned  around  once,  the  point  will 
have  traveled  a  distance  equal  to  the  circumference  of  the  pulley,  and  since 
the  pulley  turns  175  times  in  one  minute,  the  point  will  travel  175  times  the 
circumference  of  the  pulley  in  one  minute.  Consequently,  the  distance 
traveled  by  the  point  in  one  minute  is  ird  X  175  =  3.1416  X  32  X  175  = 
17592.96  in.  =  1466.08  ft.,  say  1466  ft.  Hence,  the  speed  of  the  pulley  is 
1466  ft.  per  min.     Ans. 

Example  2. — Suppose  the  speed  of  a  belt  is  3160  feet  per  minute  and  that 
it  drives  a  pulley  that  makes  330  revolutions  per  minute;  what  is  the  diam- 
eter of  the  pulley? 

Solution. — The  speed  of  a  point  on  the  circumference  of  the  pulley  is 
the  same  as  the  speed  of  the  belt,  assuming  that  there  is  no  slipping  of  the 
belt.     Consequently,  as  shown  in  example  (1), 

xrf  X  330  =3160  ft.  =  37920  in.,  or 

37920 
d  =  -^Tjf.—   =  36.5766  =  36f  J  in.,  very  nearly.     A?is. 

00\J7T 

Example  3. — The  piston  of  a  steam  engine  has  a  diameter  of  16  in.; 
what  is  the  area  of  the  piston  surface  touched  by  the  steam? 

Solution. — The  area  touched  by  the  steam  is  evidently  the  area  of  a 
circle  having  a  diameter  of  16  in.     By  formula  (4),  the  area  is 

A  =  .7854  X  162  =  201.0624,  say  201  sq.  in.     Ans. 

Example  4. — It  is  desired  to  bore  a  hole  that  shall  have  an  area  of  10  sq. 
in. ;  what  must  be  the  diameter  of  the  hole? 

Solution. — Either  formula  (4)  or  (8)  may  be  used,  but  formula  (8)  is 
rather  easier  to  apply;  using  it,  therefore, 

d  =  1.1284\/Iu=  3.5683,  say  3.568  in.     Ans. 

Example  5. — The  circumference  of  a  flywheel  was  measured  with  a 
tape  line  and  found  to  be  35  ft.  10>£  in.  What  is  its  diameter  to  the  nearest 
one-eighth  inch? 

Solution.— Either  formula  (1)  or  (5)  may  be  used.  If  (1)  be  used,  it 
will  be  necessary  to  divide  the  circumference  by  3.1416,  while  if  (5)  be  used, 
the  circumference  may  be  multiplied  by  .31831.     Since  most  computers 


§2 


MENSURATION  OF  PLANE  FIGURES 


79 


would  rather  multiply  than  divide,  use  formula  (5).  Reducing  the  feet  to 
inches,  35  ft.  10.25  in.  =  430.25  in.     Then, 

d  =  .31831  X  430.25  =  136.953  in.  =11  ft.  5  in.  to  the  nearest  one- 
eighth  inch.     Ans. 

Example  6. — When  no  ambiguity  (.confusion)  is  likely,  a  circle  may  be 
designated  by  referring  to  its  centers  only.  Fig.  55  shows  two  pulleys, 
O  and  0',  driven  by  a  belt.  Suppose  the  diameter  of  O  is  48  in.  and  it  makes 
220  revolutions  per  minute.  It  is  desired  to  have  pulley  O'  make  450 
revolutions  per  minute;  what  must  be  the  diameter  of  O'l 

Solution. — Let  N  and  n  be  the  number  of  revolutions  per  minute  made 
by  O  and  0',  respectively,  O  being  the  larger  and  O'  the  smaller  pulley; 
let  D  and  d  be  the  respective  diameters  of  O  and  O'.     The  speed  of  the  belt 


Fig.  65. 


in  feet  per  minute  is  equal  to  the  circumference  of  O  in  feet  multiplied  by 
the  number  of  times  it  turns  in  one  minute;  hence,  if  the  diameters  of  the 

■kDN 
pulleys  are  given  in  inches,  the  speed  of  the  belt  is     -.^  .     The  speed  of  the 

belt  is  also  equal  to  the  circumference  in  feet  of  O'  multiplied  by  the  number 


rdn 


Therefore, 


DN 
12 


of  times  it  turns  in  one  minute;  that  is,  it  equals  -y~ 

=  -y=-.     Dividing  both  members  of  this  equation  by  j^, 

DN  =  dn  (9) 

In  words,  the  product  of  the  diameter  and  number  of  revolutions  of  one 
pulley  is  equal  to  the  product  of  the  diameter  and  number  of  revolutions 
made  in  the  same  time  by  the  other  pulley.  Provided  the  unit  used  to 
measure  D  and  d  is  the  same,  it  is  immaterial  what  unit  is  used;  that  is, 
the  diameters  may  be  stated  in  feet,  inches,  millimeters,  etc. 
Applying  the  formula  to  the  present  case, 

48  X  220  =  d  X  450 
48  X  220 


from  which  d  = 


450 


22.41  in.  =  23H  in.  very  nearly. 


The  diameter  of  d  must  be  in  inches  because  the  diameter  of  D  is  in 
inches,  and  both  diameters  must  be  measured  in  the  same  unit.  Formula 
(9)  is  very  important  in  connection  with  calculations  pertaining  to  pulleys 
and  belts,  and  should  be  carefully  memorized. 


80  ELEMENTARY  APPLIED  MATHEMATICS  §2 

90.  Length  of  Open  Belt. — When  the  belt  passes  over  the 
pulleys  without  crossing  the. line  O'O  joining  the  centers,  as  in 
Fig.  55,  it  is  called  an  open  belt.  Knowing  the  distance  O'O 
between  the  centers  and  the  diameters  of  the  pulleys,  it  is  fre- 
quent ly  desired  to  know  the  length  of  the  belt.  There  is  no 
simple,  exact  formula  that  will  give  the  length  of  the  belt,  but  the 
formula  given  below  is  sufficiently  exact  for  all  practical  purposes. 
The  lines  A  'A  and  C'C  are  tangent  to  the  pulleys  (circles)  O 
and  0'.  Drawing  the  radii  OA  and  OC,  A  and  C  are  the  points 
of  tangency.  Drawing  the  radii  0' A'  and  O'C ,  A'  and  C  are 
points  of  tangency.  Draw  diameters  BD  and  B'D'  perpendicu- 
lar to  O'O;  then  the  angles  AOB  and  COD  are  equal,  since  the 
tangents  A  'A  and  C'C  must  intersect  in  some  point,  say  P  (not 
shown  here),  and  by  (13)  of  Art.  86,  arc  AE  =  arc  CE;  but  arc 
AB  =  90°  -  arc  AE  =  90°  -  arc  EC  =  DC,  and  AOB  =  COD. 
Since,  O'B'  is  parallel  to  OB  and  O'A'  is  parallel  to  OA,  B'O'A' 
=  BOA.  For  the  same  reason,  C'O'D'  =  COD,  and  all  four 
angles  are  equal.  The  radii  of  the  arcs  B'A'  and  BA  are  not 
equal,  and  the  length  of  the  arc  AB  is  not  equal  to  the  length 
of  the  arc  A'B'.  Let  L  =  the  length  of  the  belt,  and  let  C  =  the 
distance  O'O  between  the  centers;  then  it  is  evident  that  the 
length  of  the  belt  is  L  =  2  X  A' A  +  semicircle  BFD  +  2  X  arc 
BA  +  semicircle  B'F'D'  -  2  X  arc  B'A'.  The  difficulty  arises 
in  finding  an  expression  for  the  lengths  of  the  arcs  BA  and  B'A'. 
Using  the  same  letters  as  before,  the  following  formula  is  suf- 
ficiently exact  for  all  practical  purposes: 

L  -  2C  +  \{D  +  d)  +  ^^  (1) 

When  using  this  formula,  C,  D,  and  d  must  all  be  measured  in 

the  same  unit.     If  it  is  desired  to  measure  L  and  C  in  feet  and 

D  and  d  in  inches,  the  formula  then  reduces  to 

(D"  —  d"^2 
L  =  2C  +  .1309(Z>"  +  d")  +  *    676C°  '  (2) 

by  substituting  D"  =*  j~  and  d"  =  t~  for  D  and  d,  respectively. 

In  formula  (2),  C  means  C  feet,  and  D"  and  d"  mean  D  inches, 
and  d  inches. 

Example. — What  length  of  belt  is  required  when  the  pulleys  have  diam- 
eters of  56  in.  and  10  in.  and  the  distance  between  the  centers  is  24  ft.  3  in.? 
Solution. — Substituting  in  formula  (2)  the  value  24.25  =  24  ft.  3  in., 

for  C;  L  =  2  X  24.25  +  .1309(56  +  16)  +  ~       ^~  =  58.04-  ft.  Ans. 


§2 


MENSURATION  OF  PLANE  FIGURES 


81 


Since  .04  ft.  =  .48  in.  say  K  «i-»  the  length  of  the  belt  may  be  taken  as  58 
ft.  M  in- 

91.  Length  of  Crossed  Belt. — When  the  belt  passes  over  the 
pulleys  so  as  to  cross  the  line  O'O  joining  the  centers,  as  in  Fig. 
56,  it  is  called  a  crossed  belt.  A  and  C  are  the  points  of  tangency 
for  the  pulley  0  and  A'  and  C"  are  the  points  of  tangency  for 
the  pulley  0'.  As  in  the  case  of  the  open  belt,  the  angles  A  OB, 
COD,  A'O'B',  and  C'O'D'  are  all  equal.  The  length  of  the  belt 
is  L  =  2  X  A' A  +  semicircle  BFD  +  2  X  arc  AB  +  semicircle 


Fig.  56. 


B'F'D'  +  2  X  arc  A'B'.  As  with  the  open  belt,  there  is  no 
simple  formula  giving  an  exact  value  for  L,  but  the  following  is 
sufficiently  exact  for  all  practical  purpose: 


2C  +  ^{D  +  d)  -\ ^ — 


(1) 


Or, 


L  =  2C  +  .1309(Z>"  +  d")  + 


(D"  +  d"Y 


(2) 


576C 

The  letters  in  these  formulas  have  the  same  values  as  in  the 
formulas  of  Art.  90.  It  will  be  noted  that  the  only  difference 
between  these  formulas  and  those  of  Art.  90  is  the  sign  of  d  in  the 
last  term. 

Example. — Using  the  same  values  as  in  the  example  of  Art.  90,  find  the 
length  of  a  crossed  belt. 

Solution. — Substituting  the  values  given  in  formula  (2), 

L  =  2  X  24.25  +  .1309(56  4-  1G)  +  g^f^  24.25  =  58"296~  ^  =  5S  "' 
4%  in.,  very  nearly.  Ans.  It  will  be  noted  that  the  length  of  the  crossed 
belt,  in  this  case,  is  4%  -  H  =  4K  in.  longer  than  the  open  belt. 

92.  Concentric  and  Eccentric  Circles.— Two  or  more  circles 
are  said  to  be  concentric  when  they  have  the  same  center.  In 
Fig.  57,  the  circles  ABC  and  A'B'C  have  the  same  center  0; 
they    are,    therefore,    concentric    circles.     The    distance    A' A 


82 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


=  B'B  =  C'C,  each  distance  being  that  part  of  the  radius  included 
between  the  circles. 

If  one  circle  lies  within  another,  but  does  not  have  the  same 
center,  the  circles  are  said  to  be  eccentric.  In  Fig.  58,  the  center 
of  the  larger  circle  is  0,  and  the  center  of  the  smaller  circle  is  0'\ 
hence,  these  circles  are  eccentric  circles. 

If,  in  Figs.  57  and  58,  the  small  circle  represents  a  hole,  the 
area  of  the  space  included  between  the  hole  and  the  large  circle 


Fig.  57. 


Fig.  58. 


is  evidently  equal  to  the  area  of  the  large  circle  less  the  area  of  the 
small  circle.  Let  D  =  diameter  of  the  large  circle,  usually  called 
the  outside  diameter,  and  let  d  =  the  diameter  of  the  small 
circle,   usually  called   the  inside  diameter;  then  area   of  large 

circle  is  ^D2  and  area  of  small  circle  is  *d2.     The  area  of  the  flat 

surface  in  either  Fig.  57  or  Fig.  58  is : 

A=\D2-  \d?  -  |(Z>2  -  #).  (i) 

If  the  radius  is  used  instead  of  the  diameter, 

A  =  ir(R2  -  r2),  (2) 

in  which  R  is  the  radius  of  the  outer  circle  and  r  is  the  radius  of 
the  inner  circle. 

Example. — A  circular  disk  has  a  hole  through  it  that  is  7%  in.  in  diam- 
eter.    The  outside  diameter  is  10K  in.     What  is  the  area  of  the  flat  surface? 

Solution. — Applying  formula  (1),  remembering  that  ^  =  .7854. 

A  =  .7854  (10.252  -  7.1252)  =  42.645  -,  say  42.64  sq.  in.    Ans. 


§2 


MENSURATION  OF  PLANE  FIGURES 


83 


EXAMPLES 

(1)  It  is  desired  to  bore  a  hole  that  shall  have  an  area  of  exactly  2  sq.  in.; 
what  must  be  its  diameter?  .  Ans.   1.596— in. 

(2)  The  diameter  of  a  rod  is  4%  in.;  what  is  its  circumference? 

Ans.  14ff  in. 

(2)  A  pulley  19  in.  in  diameter  and  making  350  revolutions  per  minute 

drives  a  pulley  8  in.  in  diameter;  how  many  revolutions  per  minute  does  the 

smaller  pulley  make?  Ans.  831^  r.p.m. 

(4)  The  outside  diameter  of  a  cross-section  of  a  cylinder  is  27  in.,  the  in- 
side diameter  is  25  in.;  what  is  the  area  of  the  cross-section? 

Ans.  81.68  +  sq.  in. 

(5)  The  circumference  of  a  shaft  is  22%  in.;  what  is  its  diameter? 

Ans.  7%  in.  nearly. 

(6)  Two  pulleys  having  diameters  of  80  in.  and  20  in.  are  22  ft.  6  in.  be- 
tween centers;  what  length  of  open  belt  is  required  to  connect  these  pulleys? 

Ans.  58  ft.  4^  in. 

(7)  Two  pulleys  that  are  9  ft.  2  in.  between  centers  are  connected  by  a 
crossed  belt;  the  diameters  of  the  pulleys  being  18  in.  and  12  in.,  what  is  the 
length  of  the  belt?  Ans.  22  ft.  b&  in. 


SECTORS  AND   SEGMENTS 

93.  Circular  Measure  of  Angles. — Instead  of  measuring  an 
angle  in  degrees,  minutes,  and  seconds,  it  may  be  measured  by 
means  of  its  arc  expressed  in 
terms  of  its  radius.  For  in- 
stance, referring  to  Fig.  59, 
angle  AOB  =  angle  DOE. 
The  number  of  degrees  in  the 
angle  DOE  is  to  the  number 
of  degrees  in  the  semicircle 
GDEH  as  arc  DFE  is  to  arc 
GFH.  Letting  v  =  angle 
DOE,  v°  :  180°  =  arc  DFE  :  xr, 
r  being  the  radius  OD  and  xr 
being  Yi  X  2irr  =  length  of  arc 
of  semicircle.     From  this  pro- 

0       180°  X  arc  DFE 
portion,    v    =  

arc  ACB 
X p" >  when  R  =  radius  OA  of  arc  ACB 

that  the  arc  DFE  =  the  radius  r,  then  the  arc  ACB  must  equal 


=  57.296°  X 


arc  DFE 


=  57.296c 


Now  suppose 


the  radius  R,  and  v°  =  57.296°  X  -  =  57.296c 

r 


X  §  =  57.296°; 


84  ELEMENTAL Y   APPLIED  MATHEMATICS  §2 

that  is,  an  arc  of  57.296°  is  equal  in  length  to  the  radius  of  the  arc. 
Taking  this  arc  as  the  unit  of  measure,  in  which  case,  it  is  called  a 

180° 
radian,  a  semicircle  is  equal  to  cj  096°  =  3-1416  =  t  radians;  a 

quadrant,  or  90°,  is  half  a  semicircle  and  is  equal  to  %v  =  ^ 

=  1.5708  radians;  any  other  angle,  as  ABC,  will  be  equal  to  the 
number  of  degrees  in  the  angle  divided  by  57.296.  An  angle 
measured  in  this  manner  is  said  to  be  measured  in  radians  or  to 
be  measured  in  circular  measure.  When  an  angle  is  expressed 
in  degrees,  it  is  said  to  be  measured  in  angular  measure. 

arc  DFE 
From  the  equation  v°  =  57.296° ,  DFE  being  any  arc 

and  r  its  radius, 

™D™  =  -4-  =  angle  DFE  in  radians; 

in  other  words,  if  the  length  of  an  arc  be  divided  by  its  radius,  the 
quotient  will  be  the  measure  of  the  angle  in  radians.  Let  0  (Greek 
letter,  pronounced  theta)  be  the  angle  in  radians,  let  I  =  length 
of  the  arc,  and  r  =  the  radius  of  the  arc,  then  the  last  equation 
may  be  written 

I  =  «  (D 

From  (1),  I  =  rd  (2) 

If  an  angle  be  expressed  in  radians,  and  it  be  desired  to  find  its 

equivalent  in  angular  measure,  let  v  be  the  angle  in  degrees; 

v° 
then,  since  i^TQ^ao  =  0> 

v°  =  57.296°0.         (3) 
If  the  angle  is  given  in  angular  measure,  and  it  is  desired  to 
express  it  in  radians, 

e  =  ^~g  =  .0174533^°  (4) 

Example  i. — The  length  of  an  arc  is  23%  in.  and  the  radius  of  the  arc 
is  32  in.;  what  is  the  angle  in  radians,  and  what  is  the  angle  in  angular 
measure? 

23  625 
Solution. — From  formula  (1),  6  =      '         =  .73828  radians.     Ans. 

From  formula  (3),  v  =  57.296  X  .73828  =  42.3005°  =  42°  i8'  1.8".  Ans. 

Example  2. — A  certain  angle  is  equal  to  21°  31'  26";  if  the  radius  of  an 
arc  having  this  angle  is  15^  in.,  what  is  the  length  of  the  arc? 

Solution. — Reducing  the  minutes  and  seconds  to  a  decimal  of  a  degree, 
21°  31'  26 '=  21.5239°.  By  formula  (4),  6  =  .0174533  X  21.5239;  by 
formula   ^2),  I  =  rd  =  15.5  X  .0174533  X  21.5239  =  5.823-   in.     Ans. 


§2  MENSURATION  OF  PLANE  FIGURES  85 

94.  Length  of  Circular  Arc.— Referring  to  formula  (1),  Art. 
93,  if  r  is  equal  to  the  unit  of  linear  measure  (1  foot,  1  inch,  etc.), 

0  =  -  =  Z;  in  other  words,  the  measure  of  an  angle  in  radians 

is  the  length  of  the  arc  to  a  radius  I.  Thus,  in  Fig.  59,  if  OD  =  OE 
=  1  inch,  and  0  is  the  circular  measure  of  the  angle  AOB  =  v, 
the  arc  DFE  =  0  inches.  Further,  if  OA  =  r,  the  length  of  the 
arc  ACB  =  r  X  0  in.  =  rd  in. 

As  previously  denned,  the  figure  OACB  is  a  sector.  By  means 
of  the  formulas  and  principles  of  Art.  93,  the  length  of  the  arc 
ACB  of  any  sector  can  be  found  when  the  radius  and  the  central 
angle  (in  either  angular  or  circular  measure)  are  known.  If  the 
chord  A B  and  the  height  CI  of  the  arc  ACB  are  given,  there  is  no 
simple,  exact  formula  for  finding  the  length  of  the  arc  ACB, 
in  such  cases,  the  angle  AOB  would  usually  be  found  by  means 
of  a  table  of  trigonometric  functions;  then,  knowing  the  angle,  the 
length  I  of  the  arc  can  be  found  the  formula 

I  =  rd  =  .0174533™  (1) 

If,  however,  a  table  of  trigonometric  functions  is  not  available 

or  if  it  is  not  desired  to  use  such  a  table,  the  following  formula  will 

give  results  sufficiently  exact  for  all  practical  purposes 

_     4(H(15  +  16<2)  (2) 

1  ~  75  +  ISOt2  +  64*6  K  } 

,  .  ,  h      height  of  arc        ,  ,,  ,. 

in  which  t  =  -  =  -ufe  ,     , and  r  =  the  radius. 

c       chord  ol  arc 

To  apply  the  formula,  first  calculate  the  value  of  t;  if  r  is  not 

given,  calculate  it  by  formula  (2),  Art.  87. 

If  the  angle  is  not  greater  than  a  right  angle,  or  90°,  the  term 

64Z6  may  be  omitted,  and  the  formula  then  becomes 

_  40r*(15  +  16t2)  _  8rt(15  +  16*2)  (s) 

1  ~      75  +  180<2  15  +  36t2 

r  a/9   1 

When  the  angle  is  equal  to  90°,  -  =  t  = %~     =  -207107'> 

hence,  if  the  value  of  t  equals  or  exceeds  .21,  use  formula  (2); 
but  if  t  is  less  than  .21,  formula  (3)  may  be  used. 

Example  1.— What  is  the  length  of  an  arc  whose  chord  is  7T5S   in.  and 

whose  height  is  2^  in.? 

7t         75  117         75Y_^  W 

Solution. — Since   t  =  2\\  -5-  71C  -    32    ~   "Jg"     "32         117    ""   234 
-  .320513  is  greater  than  .21,  use  formula  (2).     By  formula  (2),  Art.  87, 

r  =  (V^+JiVJ)l  =  4.02375  in. 
8  X  2J4 


86  ELEMENTARY  APPLIED  MATHEMATICS  §2 

Substituting  in  formula  (2), 

.       40  X  4.02375  X  .320513(15  +  16  X  .3205132) 

75  +  180  X  .3205132  +  64  X  .3205136  "' 

The  correct  value  of  I  to  5  significant  figures  is  9.1748  +  in.;  hence,  the 
value  as  calculated  is  sufficiently  exact  for  practical  purposes. 

Example  2. — If  the  radius  of  the  arc  is  54  in.  and  the  height  of  the  arc  is 
8^2  in.,  what  is  the  length  of  the  arc  and  what  is  the  central  angle? 

Solution. — Referring  to  Fig.  59,  suppose  ACB  is  the  given  arc,  and  let 

OC  be  perpendicular  to  the  chord  AB;  then  AI  =  ~  and  CI  =  h.     In  the 

right  triangle  AIO,  01  =  r  -  h  and  OA  =  r.     Therefore,  s  =  "\/r2  —  (r  -  h)2 

=   V542  -  (54~^8.5)"2  =  29.0818  in.,    and   c   =  29.0818  X  2   =  58.1636 

8  5 
in.     Consequently,  t  =  5g  '        =  .14614.     Since  t  is  less  than  .21,  formula 

(3)  may  be  used,  and 

.       8  X  54  X  .14614(15  +  16  X  .146142)       _,   .„  . 

l=~  15  +  36  X  .146142  -  =  61.422  in.     Ans. 

The  correct  value  of  I  to  5  significant  figures  is  61.422  in. 

From  formulas  (1)  and  (3),  Art.  93,  d  =  -.  and  v  =  57.2960  =  57.296- 

r  r 

61.422 
=   57.296  X  — jrp  =65.171°-  =  65°  10'  15.6".     Ans. 

The  exact  value  is  65°  10'  14.6";  hence,  the  error  is  only  1  second,  which  is 
too  small  to  be  considered  in  ordinary  practice. 

If  the  angle  is  large,  that  is,  if  it  is  greater  than  90°,  and  great 
accuracy  is  desired,  it  will  be  better  to  find  the  ratio  of  the  chord 
and  height  of  half  the  arc,  which  may  be  designated  by  t'\  then 
substituting  t'  for  t  in  formula  (3),  the  length  of  half  the  arc  will 
be  found  very  closely.     By  means  of  formula  (1),  Art.  87,  and 

formulas  (1)  and  (2),  Art.  88,  the  value  of  t'  =  -  for  half  the  arc 
can  be  shown  to  be 

,  =  v^         (4) 

Or,  ,  =  vW^ie  (5) 

Having  found  t' ,  the  length  of  the  whole  arc  will  then  be  given 
by    the    formula, 

16rf  (15  +  16f") 

15  +  36£'2  (b) 

Referring  to  the  first  of  the  two  preceding  examples, 

,  WMJIJImEzM  =    X4650- 
4X2»1  '14b5,) 

Substituting  this  value  of  t'  in  formula  (6),  /  =  9.1749  +  in. 


§2  MENSURATION  OF  PLANE  FIGURES  87 

95.  Area  of  Sector  and  Segment. — The  area  of  any  sector  of  a 
circle  is  equal  to  one-half  the  product  of  its  radius  and  arc.  Let 
I  =  the    length    of    the    arc    and    r  =  the    radius,    then 

A   =  V2rl  (1) 

It  will  be  noted  that  this  formula  is  the  same  as  for  the  area  of  a 
triangle    when    I  =  the    base    and    r  =  the    altitude. 
Since    I  =  rd,    substitute    rB    for    I    in    (1),    and 

A  =  HrH  (2) 

that  is,  the  area  of  a  sector  is  equal  to  one-half  the  product  of  the 
angle  in  radians  and  the  square  of  the  radius. 

Referring  to  Fig.  59,  the  segment  ACBA  =  sector  OACB 
-  triangle  OAB.  Area  of  triangle  OAB  =  y2X  AB  XOI;b\itAB 
is  the  chord  of  the  arc  ACB  =  c,  and  01  =  radius  OC  -  minus 
height  of  arc  =  r  —  h.  Consequently,  area  of  segment  is  A 
=  lArl  -  %c(r  -  h),  or 

.        rl  —  c{r  —  h)       r26  -  c{r  -  h)  ,Q, 

A  =  ^ = 2~  (<i) 

Another  formula  for  finding  the  area  of  a  segment  (approxi- 
mately) is  the  following,  in  which  D  is  the  diameter; 

a  -  $ Vf  -  -608    (4) 

This  formula  will  give  results  sufficiently  accurate  for  most 
practical  purposes,  and  is  rather  easier  to  apply  than  (3),  when 
it  is  necessary  to  calculate  both  the  highest  h  and  the  angle  0 
of  the  arc. 

Example. — Fig.  60  represents  a  round  tank  having  an  inside  diameter 
of  60  in.  The  tank  lies  on  a  flat  surface  and  is  filled  with  water  to  a  depth 
of  42  in.  If  the  ends  of  the  tank  are  flat  and  the  tank  is  12  ft.  long,  how 
many  gallons  of  water  are  in  the  tank? 

Solution. — The  volume  of  the  water  is  equal  to  the  area  of  the  segment 
AGB  multiplied  by  the  length  of  the  tank.  To  find  the  length  of  the  arc 
AGB,  calculate  the  length  of  the  arc  AFB  and  subtract  it  from  the  circum- 
ference of  the  circle  AFBG.  The  height  of  the  arc  AFB  is  h  =  60  -  42 
=  18  in.  To  find  the  chord  AB,  apply  principle  (1),  Art.  87,  and  f^)2 
=  FH  X  HG  =  18  X  42  =  756;  from  which,  c  =  V3024  =  54.991  in.  Then, 
since  -  =  ft^t  =  -327326,  use  formula  (2),  Art.  94,  and 

C  o4   Mr/1 

.       40  X  30  X  .327326(15  +  16  X  .327326*) 
'  =  ~WTT80  X  .327326'  +  64  X  .327326s         UJO/,i     ' 
The  circumference  of  the  circle  is  3.1416  X  60  =  188.496  in.,   and   the 
length  of  the  arc  AGB  is  188.496  -  69.573  =  118.923  in.     The  area  of  the 


88 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


segment  AGB  is  evidently  equal  to  the  area  of  the  sector  AGBO  plus  the 
area  of  the  triangle  AGB;  area  of  sector  =  ^  X  30  X  118.  923  =  1783.845 
sq.  in.;  area  of  triangle  =  HX  54.991  X  (30  -  18)  =  329.946  sq.  in.;  and 
area  of  segment  =  1783.845  +329.946  -  2113.791  sq.  in.  The  length 
of  the  tank  is  12  ft.  =  144  in.;  hence,  the  volume  of  the  water  is  2113.791 
X  144  =  304385.9  cu.  in.  Since  one  gallon  contains  231  cu.  in.,  the  number 
of  gallons  in  the  tank  is  304385.9  *  231  =  1317.7 -,  say  1318  gallons.  Ans. 


Fig.  GO. 


The  area  of  the  segment  might  have  been  found  by  finding  the  area  of  the 
small  segment  AFB  and  subtracting  it  from  the  area  of  the  circle.  The 
flat  surface  ABCD  represents  the  water  level. 

96.  Area  of  Fillet. — When  two  solids  intersect  so  as  to  form  a 
square  corner,  as  shown  in  Fig.  61,  the  strength  of  the  piece  can 
be  greatly  increased  by  rounding  the  corner.     This  is  usually 

done  by  striking  an  arc  of 
a  circle  having  a  small 
radius.  The  curved  part 
thus  added  is  called  a  fillet. 
While  fillets  add  somewhat 
to  the  weight  of  the  piece, 
it  is  customary  to  neglect 
this  in  the  case  of  heavy 
castings;  but  if  an  accurate  estimate  of  the  weight  is  desired, 
it  is  necessary  to  include  the  weight  of  the  fillets.  Referring  to 
Fig.  61,  A CB  is  a  fillet,  and  it  is  assumed  that  the  sides  AC  and 
BC  form  a  right  angle.  Letting  OA  =  OB  =  r,  the  area  of  the 
fillet  is  equal  to  the  area  of  the  square  AOBC  -  area  of  the 
quadrant  AOB;  hence,  A  =  r*  -  i^rr2  =  r2(l  -  .7854),  or 

A  =  .2146r2. 


Fia.  61. 


§2  MENSURATION  OF  PLANE  FIGURES  89 

Example.— If  the  radius  of  a  fillet  is  Y±  in.,  what  is  the  area  of  the  fillet? 

Solution.— Applying  the  formula,  A  =  .2146  X  (K)2  =  .01341  sq.  in. 

Ans. 

It  will  be  noted  that  the  area  is  quite  small;  and  since  the  radius 
is  never  very  large,  it  is  sufficiently  exact  for  practical  purposes 
to  take  the  value  of  A  as  one-fifth  the  square  of  the  radius. 


EXAMPLES 

(1)  If  the  length  of  an  arc  is  10.26  in.  and  its  radius  is  33.14  in.,  what  is 
the  angle  in  radians  and  also  in  angular  measure?      .        j  .30960— radians. 

Am-\  17°  44'  19". 

(2)  If  a  certain  angle  measures  31°  12'  27",  what  is  the  length  of  the 
arc  having  this  angle,  the  radius  of  the  arc  being  19.32  in.?  Ans.  10.523  in. 

(3)  The  chord  of  an  arc  is  14%  in.,  the  height  of  the  arc  is  1%  in.,  what  is 
the  length  of  the  arc  and  what  is  the  angle  in  radians  and  degrees? 

[  I     =    15.378  in. 
Ans.    I  6     =    .99586  radian. 
I  v°  =  57°  3'  32". 

(4)  Referring  to  the  example  of  Art.  95,  how  many  gallons  are  in  the 
tank  when  the  depth  of  the  water  is  54  in.?  Ans.  1671  gallons. 

(5)  A  fillet  has  a  radius  of  %6  in.;  what  is  the  area  of  the  fillet? 

Ans.  .021  —  sq.  in. 


INSCRIBED  AND  CIRCUMSCRIBED  POLYGONS 

97.  An  inscribed  polygon  is  one  whose  vertexes  lie  on  the 
circumference  of  a  circle.  In  Fig.  62  the  vertexes  of  the  poly- 
gon ABCDEF  lie  on  the  circumference  of  the  circle  ABCDEF, 
and  it  is,  therefore,  an  inscribed  polygon.  A  circumsciibed 
polygon  is  one  whose  sides  are  tangent  to  a  circle.  In  Fig,  62. 
the  sides  of  the  polygon  ABCDEF  are  all  tangent  to  the  circle 
A'B'C'D'E'F',  and  it  is,  therefore,  a  circumscribed  polygon  with 
reference  to  the  circle  A'B'C'D'E'F'. 

If  the  polygon  is  a  regular  polygon,  it  may  always  be  inscribed 
in  a  circle,  and  it  may  also  be  circumscribed  about  a  circle.  The 
polygon  in  Fig.  62  is  a  regular  hexagon,  and  it  is  a  property  of  the 
regular  hexagon  that  the  sides  are  always  equal  in  length  to  the 
radius  of  the  circle  in  which  the  hexagon  is  inscribed;  thus  AB  = 
BC  =  OC.  Consequently,  a  regular  hexagon  may  be  drawn  by 
describing  a  circle  and  then  spacing  off  on  its  circumference 
chords  equal  in  length  to  the  radius. 

The  center  of  the  circles  within  which  the  regular  polygon  is 
inscribed  and  about  which  it  is  circumscribed  is  the  geometrical 


90 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


center  of  the  polygon,  and  lines  drawn  from  the  center  to  the 
points  of  tangency,  as  OA',  OB',  etc.  are  apothems  of  the 
circumscribed  polygon.  Let  a  =  the  apothem  and  r  =  the 
radius  of  the  circle   within  which  the   regular  polygon  is  in- 

S  5 

scribed;  then,  if  s  =  a  side,  ~  =  half  a  side=  B'B,  and  U)2^  f2 
—  a2,  from  which 

s  =  4\/r2  —  a2. 

Given  the  radius  r  and  the  length  5  of  the  side,  describe  a  circle 
with  the  radius  r,  and  then  space  off  on  the  circumference  the 
length  s.  The  value  of  a  for  polygons  most  commonly  used  may 
be  obtained  from  the  table  given  in  Art.  81,  in  which  a  is  a 
multiple  of  s. 


Fig.  62. 


Fio.  63. 


Example. — It  is  desired  to  inscribe  a  regular  polygon  of  11  sides  in  a  circle 
having  a  diameter  of  2]/2  in.  Show  how  this  can  be  done  by  means  of  the 
table  in  Art.  81. 

SOLUTION. — The  apothems  given  in  the  table  are  multiples  of  the  length 
of  the  side;  hence,  if  the  side  be  taken  as  1  inch,  the  apothem  for  a  regular 
polygon  of  11  sides  will  be  1.7028  in.,  say  1.7  in.  Draw  two  lines  AB  and 
OC  at  right  angles  to  each  other,  Fig.  63.  Make  OC  equal  to  1.7  in.,  the 
apothem,  and  CB  =  CA  =  \s  =  \  in.,  in  this  case  (taking  s  as  1  in.), 
and  draw  OA  and  OB.  With  O  as  a  center  and  a  radius  equal  to  the  radius 
of  the  given  circle  =  \  X  2\  =  \\  in.,  describe  a  circle;  join  the  points 
E  and  F,  the  points  of  intersection  of  OA  and  OB  with  the  circle,  by  the  line 
BF,  and  KF  will  be  one  of  the  sides  of  the  inscribed  11-sided  polygon,  and 
may  be  spaced  off  around  the  given  circle. 


§2 


MENSURATION  OF  PLANE  FIGURES 


91 


If  it  is  desired  to  calculate  the  length  of  the  side  EF,  first 
find  the  length  of  OB.     In  the  right  triangle  OCB,  CB  =  .5, 

=  v& 

EF       OF         s         1.25 


and    OC  =  1.7028;    hence,    OB  --     V-52  +  1.70282  =■■  1.7747 
The  triangles  AOB  and  EOF  are  similar;  whence  the  pro- 


portion -£g  =  TTg'  or  7  =  i  7747>  and  «  =  -70434  in. 

If  the  line  AB  be  taken  as  some  length  other  than  1  inch,  it 
must  be  multiplied  by  the  value  of  the  apothem  as  given  in  the 
table  to  find  the  length  of  OC. 


THE    ELLIPSE 

98.  The  ellipse  is  a  plain  figure  so  constructed  that  the  sum 
of  the  distances  from  any  point  on  the  curve  to  two  fixed  points 
is  constant;  thus,  referring  to  Fig.  64,  F  and  F'  are  the  two  fixed 
points,  and  CF  +  CF'  =  PF  4-  PF'  =  QF  4-  QF'  =  etc.,  and  the 


closed  curve  is  an  ellipse.  The  longest  line  that  can  be  drawn  in 
the  figure  is  the  line  A  B,  which  passes  through  the  two  fixed  points 
F  and  F';  it  is  called  the  major  axis.  From  the  center  0  of  the 
major  axis,  draw  CD  at  right  angles  to  AB;  then  CD  is  the  short- 
est line  that  can  be  drawn  in  the  ellipse,  and  is  called  the  minor 
axis.  The  points  A  and  B  at  the  extremities  of  the  major  axis 
are  called  the  vertexes  of  the  ellipse;  the  point  0  is  called  the 
center,  and  the  fixed  points  F  and  F'  are  called  the  foci,  either 


92  ELEMENTARY    \PPLIED  MATHEMATICS  §2 

point  being  called  a  focus  of  the  ellipse.  The  vertexes,  center, 
and  foci  all  lie  on  the  major  axis,  and  the  distance  AF  must 
equal  the  distance  F'B\  hence,  for  the  point  A,  FA  +  F'A 
=  F'A  +  F'B  =  AB;  and  the  sum  of  the  distances  from  any  point 
on  the  curve  to  the  foci  is  equal  to  the  major  axis.  For  the  point 
C,  CF  =  CF'  =  %AB  =  OA  =  OB. 

In  practice,  an  ellipse  is  usually  specified  by  giving  the  lengths 
of  the  major  and  minor  axes,  which  are  commonly  called  the  long 
and  short  diameters.  An  ellipse  may  be  drawn  mechanically 
in  the  following  manner:  Lay  off  the  long  diameter  (major 
axis),  and  bisect  it,  thus  locating  the  center  0;  through  0,  draw 
a  perpendicular  CD,  and  lay  off  OC  =  OD  =  one-half  the 
short  diameter  (minor  axis);  then,  with  C  (or  D)  as  a  center 
and  with  a  radius  equal  to  one-half  the  long  diameter,  draw  short 
arcs  cutting  AB  in  F  and  F',  thus  locating  the  foci.  Stick  pins 
in  the  paper  at  F  and  F'  and  also  at  C;  tie  one  end  of  a  piece  of 
thread  to  one  of  the  pins  F  or  F',  and  pass  the  thread  around  the 
other  two  pins,  drawing  it  taut  and  passing  it  several  times  around 
the  pin  at  the  other  focus.  Now  pull  out  the  pin  at  C,  and  with 
a  pencil  held  perpendicular  to  the  plane  of  the  paper  and  pressing 
against  the  thread  (but  not  hard  enough  to  stretch  it),  move  it  so 
that  it  keeps  the  thread  tight,  thus  describing  one-half  of  the  el- 
lipse, say  the  upper  half  ACPB;  then  bring  the  thread  to  the 
other  side  of  the  pins,  describe  the  other  half  of  the  ellipse  or 
ADQB. 

99.  Circumference  and  Area  of  Ellipse. — In  works  on  mathe- 
matics, it  is  customary  to  denote  one-half  the  major  axis  by  the 
letter  a  and  one-half  the  minor  axis  by  the  letter  b;  thus,  in  Fig. 
64,  a  =  OA  =  OB,  and  b  =  OC  =  OD.  The  area  of  an  ellipse 
is 

A  =  irab  (1) 

There  is  no  exact  formula  giving  the  circumference  (periphery) 

of  an  ellipse,  but  the  following  formula,  in   which  r  = r 

a  -f-  o 
and  p  =  the  circumference  (periphery),  gives  results  sufficiently 
exact  for  all  practical  purposes: 

,     ,   , ,  64  -  3r4 
V  -  »(«  +  b)  ^--^  (2) 

Example. — The  long  and  short  diameters  of  an  ellipse  are  16J4  in.  and 
A)4  in.  respectively;  what  is  the  area  of  the  ellipse?  what  is  the  circumference. 


§2  MENSURATION   OF  PLANE  FIGURES 

Solution. — The  area,  by  formula  (1),  is 


93 


A  =*ab  =  3.1416  X~z5X~ 


3.1416 


X  16.25  X  4.5 
A  ns. 


=  .7854  X  16.25  X  4.5  =  57.432+  sq.  in. 

The  circumference  (periphery),  by  formula  (2),  is 

_-.,-  /16.25  .  4.5X64  -  3  X  .56627'        ,_  -..  . 
v  -  3.1416  (-j-  +  T  j  ^  _  16  x  j66627,  -  35.264  in.     An,. 

To  apply  formula   (2),  first  calculate  the  value  of  r;  thus,  r  = 


o  -  b 
a  +  b 


2<z  -  2b       16.25  -  4.5 


.56627-.     Then,  r2  =  .320662  -;    r4  =  (r2)J 


~  2a  +  26        16.25  +  4.5 

=  .3206622  =  .102824  +  ;  the  remainder  of  the  work  is  evident. 

It  may  be  remarked  that  the  major  axis  divides  the  ellipse 

into  two  equal  parts;  the  minor  axis  also  divides  the  ellipse  into 

two  equal  parts. 

AREA  OF  ANY  PLANE  FIGURE 

100.  If  the  figure  can  be  divided  into  elementary  plane  figures 
(i.e.  triangles,  rectangles,  circles,  segments,  etc.)  the  area  of  each 
of  the  elementary  fig- 
ures may  be  calculated, 
and  the  sum  will  be  the 
area  of  the  entire  figure. 
Referring  to  Fig.  65,  the 
dotted  lines  show  that 
the  figure  may  be 
divided  into  three  trape- 
zoids two  of  them  equal 
- — one  rectangle,  a  seg- 
ment of  a  circle,  and 
two  equal  fillets.  From 
the  dimensions  marked 
on  the  drawing,  the 
areas  of  all  these  ele- 
mentary figures  may  be 
found,  and  their  sum 
will  be  the  area  of  the 
entire  figure.  The  work 
is  as  follows: 

There  are  two  trapezoids  of  the  same  size  as  abdc;  the  length  of 
the  side  ac  is  2|  +  f  =  3",  and  the  area  of  the  two  trapezoids  is 
3  +  2^ 


Fio.  65. 


A   = 


X  3f  X  2  =  17.72  sq.  in. 


94 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


The  length  of  the  rectangle  is  2\  +  f  +  4|  =  7£",  and  its 
area  is 

A  =  7i  X  If  =  12.47  sq.  in. 
The  length  of  the  side  ih  of  the  trapezoid  kjhi  is  1|  +  If  +  l\ 


=  4f"  and  its  area  is 


4f  +  2i 


X  3  =  10.5  sq.  in. 


The   area   of   the   fillets   is   2XJX  (f)2  =  .06   sq.  in. 

Concerning  the  segment,  the  radius  and  the  chord  are  known  and 
it  is  necessary  to  calculate  the  height.  Using  formula  (3), 
Art.  87,  h  =  r  -  y2\/±r*  -  c2  =  1.5  -  HV*  X  1.5s  -  2.252 
=  .508". 

Substituting  this  value  of  h  in  formula  (4),  Art.  95, 

4  X  5082     /2  X  1.5 


3 

area 


4 


.508 


.608  =  .79  sq.  in. 


is 


17.72  +  12.47  +  10.5  +  0.06  +  0.79 


The     entire 
=  41.54  sq.  in. 

101.   If  the  figure  is  of  such  a  nature  that  it  cannot  be  divided 
into  simple  elementary  figures,  proceed  as  in  Fig.  66.     Draw  a 


Fig.  66. 


line  AB,  either  through  the  figure  or  outside  of  it,  as  in  Fig.  66; 
then  draw  lines  CB  and  DA  tangent  to  the  extreme  ends  of  the 
given  outline  and  perpendicular  to  the  line  AB.  The  intersec- 
tion of  the  tangents  with  AB  locate  the  points  A  and  B.  Divide 
AB  into  any  convenient  even  number  of  equal  parts,  in  the  present 
case  8,  and  through  the  points  of  division,  erect  perpendiculars 
(called  ordinates)  to  AB;  these  intersect  the  figure  in  the  lines 


§2  MENSURATION  OF  PLANE  FIGURES  95 

a'a,  b'b,  c'c,  etc.,  which  are  all  parallel  to  one  another,  and  are 
equally  distant  apart.  If  n  =  the  number  of  parts  into  which 
AB  has  been  divided,  and  h  =  the  distance  between  two  con- 

AB 

secutive    ordinates,    measured    parallel    to    AB,    h  = 

If,  now,  another  series  of  ordinates  are  drawn  midway  between 
those  previously  drawn,  here  indicated  by  the  dotted  lines,  and 
the  lengths  of  these  dotted  ordinates  are  measured  and  added, 
the  sum  so  obtained  multiplied  by  h  will  be  the  approximate  area 
of  the  figure.  This  method  is  called  the  trapezoidal  rule.  The 
greater  the  number  of  parts  into  which  AB  is  divided,  that  is, 
the  smaller  the  value  of  h,  the  more  accurate  will  be  the  result 
obtained    for   the    area. 

102.  Another  rule  that  is  more  accurate  than  the  method  just 
described  is  what  is  known  as  Simpson's  rule.  Having  drawn 
the  ordinates  dividing  the  figure  into  n  equal  parts  (n  being  an 
even  number),  designate  the  ordinates  by  y0,  yi,  y%,  etc.,  as  shown 
in  the  figure.  Then  letting  yQ  and  y„  be  the  end  ordinates,  the 
area    by    Simpson's    rule    is 

A  =  ^  [y0  4-  yn  4-  Myi  +  y3  +  y&  +  etc.)  +  2(y2  +  yi  4-  y6  +  etc.)] 

Expressed  in  words,  Simpson's  rule  is: 

Add  the  end  ordinates,  jour  times  the  odd  ordinates,  and  2  times 
the  remaining  even  ordinates,  and  multiply  this  sum  by  one-third 
the  distance  between  the  ordinates. 

Applying  both  methods  to  the  outline  given  in  Fig.  66,  the 
lengths  of  the  dotted  ordinates,  beginning  with  the  left  and  pro- 
ceeding in  order  to  the  right,  are  0.97",  1.52",  1.51",  1.04",  0.57", 
0.67",  0.85",  and  0.75";  their  sum  is  7.88".  The  number  of 
equal    parts    is    8  =  n,    and     I  =  AB  =  3.02".     Therefore,    h 

3.02         ,     .       3.02      ___      3.02X7.88      on„         .       , 
=  -g-,   and  A  =  -g—  X  7.88  = g =  2.97  sq.  in.,  by 

the  trapezoidal  rule. 

Applying  Simpson's  rule,  y0  and  ys  are  both  equal  to  0;  yi 
=  1.34",  y3  =  1.34",  yb  =  0.64",  y,  =  0.85",  and  the  sum  of  these 
odd  ordinates  is  4.17";  y2  =  1.56",  y4  =  0.67",  y6  =  0.83", 
and  the  sum  of  these  even  ordinates  is  3.06".     Therefore,  by 

j  o  no  1 

the    rule,    since    „  =  ■  '-~-  X  «,  =  .12583, 

A  =  .12583  (0  +  0  +  4  X  4.17  +  2  X  3.06)  =  2.87  sq.  in. 


96  ELEMENTARY  APPLIED  MATHEMATICS  §2 

103.  It  was  stated  in  Art.,  101  that  the  greater  the  number  of 
ports  into  which  AB  is  divided  the  greater  the  accuracy  of  the 
result,  and  this  is  true  whichever  method  is  used.  If  AB  be 
divided  into  16  equal  parts  ins-tead  of  8,  the  area  will  be  2.954 
sq.  in.  by  the  trapezoidal  rule  and  2.890  by  Simpson's  rule.  For 
32  equal  divisions,  Simpson's  rule  gives  2.919  sq.  in.  Tabulat- 
ing   these    results, 

Trapezoidal  rule  (8  parts) 2.975  sq.  in.  Error,  +0.056 

Trapezoidal  rule  (16  parts) 2.954  sq.  in.  Error,  +0.035 

Simpson's  rule  (8  parts) 2.869  sq.  in.  Error,  —0.050 

Simpson's  rule  (16  parts) 2.890  sq.  in.  Error,  -0.029 

Simpson's  rule  (32  parts) 2.919  sq.  in.  Error,  0 

Assuming  that  the  result  obtained  by  Simpson's  rule  for  32 
parts  is  correct  to  all  four  figures,  it  will  be  observed  that  the 
error  when  using  the  trapezoidal  rule  is  somewhat  greater 
than  when  using  Simpson 's  rule.  In  practice,  it  is  always  advis- 
able to  divide  AB  into  at  least  10  parts;  this  not  only  increases 
the  accuracy,  but  it  makes  n  a  very  convenient  number  to  divide 
by  when  finding  h.  If  particularly  accurate  results  are  desired, 
it  is  best  to  make  n  =  20,  in  which  case,  the  result  will  be  as 
accurate    as    the    limits    of    measurement    permit. 


ELEMENTARY  APPLIED 
MATHEMATICS 

(PART  2) 


EXAMINATION  QUESTIONS 

(1)  The  length  of  the  diagonal  of  a  square  is  5y  inches;  what  is 
(a)  the  length  of  one  of  the  sides,  and  (b)  what  is  the  area? 

(a)     3.7123  in. 


1  (6)   13.781  sq.  in. 

(2)  The  lengths  of  the  sides  of  a  triangle  are  29.1  ft.,  21.8  ft., 
and  36.5  ft.;  what  is  the  area  of  the  triangle? 

Ans.  317.18  sq.  ft. 

(3)  The  side  of  a  regular  dodecagon  measures  4§  in.;  what  is 
the  area  of  the  polygon?     •  Ans.  266.08  sq.  in. 

(4)  If  (a)  the  diameter  of  a  circle  is  35f  in.,  what  is  its  area? 
(6)  What  must  be  the  diameter  of  a  circle  that  will  enclose  an 
area  of  1800  sq.  in.?  a         [  (a)  1003.8  sq.  in. 

AnS-    I  (6)       47.873  in. 

(5)  The  length  of  the  chord  of  a  circular  arc  is  46|  in.,  the 
height  of  the  arc  is  10|  in.,  what  is  (a)  the  radius?  (6)  the  angle 
in  radians?  (c)  the  angle  in  degrees?     First  find  length  of  arc. 

r  (a)  31.636  in. 
Ans.    \  (6)   1.6570  radians. 
I  (c)  94°  56'  27" 

(6)  Referring  to  the  last  example,  what  is  (a)  the  chord  of 
half  the  arc?  (6)  what  is  the  height  of  half  the  arc? 

(a)  25.466  in. 


'  (6)  2.6756  in. 

(7)  The  diameters,  of  two  belt  pulleys  are  64  in.  and  14  in. 
and  the  distance  between  the  centers  is  18  ft.  9  in.  (a)  What 
length  of  open  belt  is  required?  (b)  what  length  of  crossed  belt? 
Give  lengths  to  nearest  16th  of  an  inch. 

Ans     /(<0  47ft.llftin. 
Am'    I  (6)  48  ft.  3A  in. 
7  '  97 


98 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


(8)  If  the  central  angle  of  a  circular  arc  is  121°  27'  36",  (a) 
what  is  the  angle  in  radians?  (b)  if  the  radius  is  241  ft.,  what  is 
the  length  of  the  arc?  ,         /  (a)  2.1199  radians. 

AnS-    1(6)  51.938  ft. 

(9)  Find  from  the  dimensions  given,  the  area  of  the  outline 
shown  in  Fig.  1.  The  dotted  lines  indicate  how  the  figure  should 
be  divided.  As  shown,  the  figure  is  divided  into  two  semi-circles, 
two  rectangles,  two  trapezoids,  two  sectors,  and  two  fillets. 

Ans.  28.261  sq.  in. 


■% 


J* 


rail. 


V* 


I 


4  rati. 


r 


* 


2 

Fig.   1. 


(10)  If  the  chord  of  a  circular  arc  is  8.5  in.  and  the  height  of 
the  arc  is  1 ,5ff  in.,  what  is  (a)  the  radius?  (6)  the  length  of  the 
arc?  (c)  the  angle  in  radians?  (d)  the  angle  in  degrees?  (e)  the 


area  of  the  sector? 


Ans. 


(a)  7.5372  in. 

(b)  9.0307  in. 

(c)  1.1981  radians. 

(d)  68°  387  56" 

(e)  34.033  sq.  in. 


§2 


EXAMINATION  QUESTIONS 


99 


Ans. 


(11)  Referring  to  the  last  question,  (a)  what  is  the  area  of  the 
segment?     (6)  Calculate  the  area  also  by  formula  (4)  of  Art.  95. 

(a)  7.578  sq.  in. 
(6)   7.575  sq.  in. 

(12)  If  the  major  and  minor  axes  of  an  ellipse  are  37  in.  and 
122  in-j  respectively,  (a)  what  is  the  area?  (b)  what  is  the  peri- 
phery of  the  ellipse?  j  (a)  363.25  sq.  in. 

Am-    1(6)     82.594  in. 

(13)  The  outline  shown  in  Fig.  2  resembles  an  indicator  dia- 
gram showing  the  variation  of  pressure  in  a  steam-engine  cylin- 
der.    Find  the  area  of  the  diagram  (a)  by  the  trapezoidal  rule, 


Fig.  2. 


and  also  by  (b)  Simpson's  rule.  The  end  ordinates  should  be 
tangent  to  the  curve  at  C  and  D  and  perpendicular  to  a  line 
parallel  to  AB.  The  value  obtained  by  dividing  the  area  by  the 
perpendicular  distance  between  the  end  ordinates  is  called  the 
■mean  ordinate,  and  is  indicated  on  the  diagram  by  the  dotted 
line  MN;  that  is,  it  is  equal  to  the  perpendicular  distance  be- 
tween MN  and  AB.  Find  (c)  value  of  the  mean  ordinate.  Divide 
diagram  into  12  equal  parts;  for  the  convenience  of  the  student, 
the  orcjinates  have  been  drawn  in  position. 

(14)  A  certain  building  lot  has  the  shape  of  a  sector  of  a  circle 
whose  radius  is  149  ft.,  the  frontage  being  117  ft.  What  is  the 
area  of  the  lot,  and  what  part  of  an  acre  is  it? 

Ans.  8716.5  sq.  ft.  =  .20010  A. 

(15)  An  opening  is  to  have  the  shape  of  an  equilateral  triangle 
and  is  required  to  have  an  area  of  16  sq.  in.;  what  is  the  length 
of  one  of  the  sides?  Ans.  6.0787  in. 


100        ELEMENTARY  APPLIED  MATHEMATICS  §2 

(16)  Referring  to  Question  7.  suppose  the  speed  of  the  belt 
is  2640  ft.  per  min.  and  there  is  no  slip;  (a)  how  many  revolutions 
per  minute  does  the  large  pulley  make?  (6)  how  man}'  does  the 
small  pulley  make?  .  f  (a)  157.56  r.p.m. 

I  (6)   720.30  r.p.m. 


ELEMENT  ABY  APPLIED 
MATHEMATICS 


(PART  3) 


MENSURATION  OF  SOLIDS 


PRISMS,  CYLINDERS,  CONES,  AND  SPHERES 


POLYEDRONS 

104.  Every  solid  has  three  dimensions — length,  breadth,  and 
thickness;  see  Art.  43.  A  simple  example  of  a  solid  is  shown  in 
Fig.  67.  Here  the  sides  aa'b'b,  bb'c'c,  cc'd'd,  and  aa'd'd  are  called 
the  faces  or  lateral  sides;  the  end  sides  abed  and  a'b'c'd'  are  called 
the  bases.  Both  ends  and  all  the  faces  are  flat,  i.e.,  they  form 
parts  of  plane  surfaces,  and  in  the  illustration,  the  ends  are 
parallel  and  the  opposite  faces  are  also  parallel. 


Fig.   67. 

The  planes  of  the  faces  intersect  in  right  lines  to  form  the 
edges  of  the  solid;  in  Fig.  67,  the  edges  formed  by  the  intersec- 
tion of  the  sides  are  a'a,  b'b,  e'e,  and  d'd.  The  planes  of  the  faces 
also  intersect  the  planes  of  the  bases  in  right  lines  to  form  the 
edges  at  the  ends  of  the  solid ;  in  Fig.  67,  the  edges  formed  by  the 

101 


102        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


intersection  of  the  faces  and  ends  are  ab,  be,  cd,  da,  a'b',  b'c', 
c'd\  and  d'a'. 

105.  Prisms. — Any  solid  whose  ends  and  sides  are  made  up  of 
plane  surfaces  is  called  a  polyedron.  If  the  polyedron  has  two 
efiual  and  parallel  bases,  it  is  called  a  prism,  and  if  the  bases  of  the 
prism  are  rectangles  or  squares,  it  is  called  a  parallelopiped. 
The  solid  shown  in  Fig.  67  is  a  prism  and  also  a  parallelopiped. 
The  bases  of  a  prism  may  be  polygons  of  any  shape — regular  or 
otherwise — but  they  must  be  parallel  and  equal. 

106.  If  the  planes  of  the  bases  are  perpendicular  to  the  planes 
of  the  sides,  the  lateral  edges  are  perpendicular  to  the  edges  of  the 
bases,  and  the  prism  is  called  a  right  prism.  Fig.  67  shows  a 
right  prism,  which  is  also  a  right  parallelopiped. 

107.  If  a  prism  be  intersected  by  a  plane  at  right  angles  to  the 
lateral  edges,  as  the  plane  MN  in  Fig.  67,  the  outline  of  the  plane 
figure  so  formed  will  be  equal  to  the  bases,  and  the  section  is 
called  a  right  section  or  a  cross  section  (usually,  the  latter). 
In  Fig.  67,  the  plane  MN  intersects  the  prism  in  the  cross  section 
a  1 6  yc  ld    ,  which  is  equal  and  parallel  to  the  base  abed. 

108.  If  a  prism  be  intersected  by  a  plane  at  right  angles,  i.e., 
perpendicular  to,  the  bases,  the  section  so  formed  is  called  a 

longitudinal  section;  in  Fig.  67,  the  plane  PQ 
intersects  the  prism  in  the  longitudinal  section 
a"b"b'"a'" ,  the  plane  PQ  being  perpendicular 
to  both  bases.  Whatever  the  shape  of  the 
prism,  every  longitudinal  section  is  a  rectangle, 
since  it  can  cut  only  two  faces  at  one  time,  or 
if  it  cuts  three  faces,  it  must  pass  through  one 
of  the  edges,  thus  forming  but  one  side  of  the 
rectangle. 

109.  The  altitude  of  a  prism  is  the  perpen- 
dicular distance  between  the  bases.  Thus,  in 
Fig.  68,  suppose  that  the  sides  are  not  at  right 
angles  to  the  bases;  then  h,  which  represents 

the  perpendicular  distance  between  the  planes  of  the  bases,  is 

the  altitude. 

110.  Referring  to  Figs.  67  and  68,  it  will  be  observed  that  the 
sides  of  any  prism  are  parallelograms,  and  when  the  prism  is  a 
righl  prism,  the  sides  are  rectangles.  Since  the  bases  are  parallel, 
all  the  lateral  edges  are  of  equal  length. 


Fig.  68. 


§2  MENSURATION  OF  SOLIDS  103 

111.  Area  and  Volume  of  Prism. — By  lateral  area  is  meant  the 
area  of  the  outside  of  the  solid  not  counting  the  ends;  in  the  case 
of  a  prism,  it  is  equal  to  the  area  of  the  faces. 

Referring  to  Fig.  67,  note  that  if  the  plane  MN  is  perpendicular 
to  one  of  the  faces,  it  is  perpendicular  to  all  of  them,  and  the  lines 
of  intersection  of  the  plane  with  the  faces  are  perpendicular  to  the 
edges;  thus,  aIvbIV  is  perpendicular  to  a' a  and  b'b,  b1  c1  is  per- 
pendicular to  b'b  and  c'c,  etc.  Hence,  these  lines  are  equal  to  the 
altitudes  of  the  parallelograms  forming  the  faces,  the  lengths 
being  all  equal  to  the  lengths  of  the  lateral  edges,  which  are  all 
equal.  Consequently,  the  lateral  area  of  a  prism  is  equivalent  to 
that  of  a  rectangle  whose  length  is  one  of  the  lateral  edges  and 
whose  altitude  is  equal  to  the  sum  of  the  lengths  of  the  lines 
formed  by  the  intersection  of  the  plane  MN  (perpendicular  to  the 
lateral  edges)  with  the  faces  of  the  prism,  and  this  latter  is  equal 
to  the  perimeter  of  the  polygon  formed  by  the  intersection  of 
the  plane  MN  with  the  faces. 

Let  A  =  the  lateral  area  of  the  prism,  I  =  length  of  a  lateral 
edge,  and  p  =  the  perimeter  of  the  polygon  formed  by  a  right 
section  of  the  prism  =  perimeter  of  either  base;  then, 

A  -  pi  (1) 

If  the  area  of  the  ends  is  also  included,  the  result  is  called  the 
entire  area.  Let  Ac  =  the  entire  area  and  let  a  =  the  area  of 
one  end;  then, 

Ac  =  A  +  2a  =  pi  +  2a  (2) 

112.  By  volume  of  a  solid  is  meant  the  number  of  cubic  units 
that  it  contains;  if  the  unit  of  measurement  is  one  inch,  then  the 
volume  is  the  number  of  cubic  inches  that  the  solid  would  occupy 
if  made  of  some  soft  material  that  could  be  formed  into  a  cube. 
The  phrase  cubical  contents  is  frequently  used  instead  of  the  word 
volume,  and  the  two  terms  have  identically  the  same  meaning. 

Let  V  =  the  volume  of  a  prism,  a  =  the  area  of  one  of  the 
bases,  and  h  =  the  altitude;  then, 

V  =  ah  (1) 

That  is,  the  volume  of  a  prism  is  equal  to  the  area  of  the  base 
multiplied  by  the  altitude. 

If  the  base  is  a  rectangle,  let  I  =  the  length  and  b  =  the 
breadth;  then, 

V  =  blh  (2) 


104         KLKMENTAKY  APPLIED  MATHEMATICS  §2 

If  the  base  is  a  square,  let  d  =  the  length  of  one  of  the  sides; 
then, 

V  =  d%  (3) 

If  the  base  is  a  square  and  the  altitude  is  equal  to  one  of  the 
sides  of  the  base, 

V  =  hz  (4) 

If  the  prism  is  a  right  prism,  the  altitude  is  equal  to  one  of  the 
lateral  edges,  and  if  it  is  also  equal  to  one  of  the  edges  of  the  base, 
the  prism  is  a  cube,  and  its  volume  is  found  by  formula  (4) . 

Example  1. — A  room  is  22  feet  long,  14  ft.  8  in.  wide,  and  9  ft.  10  in. 
high;  how  many  cubic  feet  of  air  does  the  room  contain? 

Solution. — The  cubical  contents  of  the  room  is,  by  formula  (2),  the  prod- 

uct  of  the  length,  width,  and  height.     14  ft.  8  in.    =  -j~   ft.  and  9  ft.  10 

118  ,       TT  Tr       nn      176       118      456896       0,_08         .,     . 

in.  =  y2  ft.     Hence,  V  =22Xj^  X^=      U4      =  3172^  cu.  ft.  Ans. 

Example  2. — A  blowpit  20'  X  10'  X  14'  is  filled  with  stock  weighing 
65  lb.  per  cubic  foot;  what  is  the  weight  of  the  stock?  If  the  stock  is 
5.7%  fiber,  what  is  the  weight  of  the  fiber? 

Solution.— The  cubical  contents  of  the  blowpit  is  20  X  10  X 14  =  2800 
cu.  ft.  Since  1  cu.  ft.  of  stock  weighs  65  lb.,  the  weight  of  the  stock  is  2800 
X  65  =  182,000  lb.     Ana. 

Since  5.7%  of  the  stock  is  fiber,  the  weight  of  the  fiber  is  182,000  X  .057 
=  10,374  lb.     Ans. 

113.  Pyramids. — A  pyramid  is  a  polyedron  having  one  base,  a 

polygon,  and  whose  sides  are  triangles  meeting  at  a  common  point 

called  the  vertex.  „  When  a  pyramid  is 

h\  referred  to  by  letters  placed  at  the  vertex 

I  j\Y\  and  at  the  corners  of  the  base,  the  letter 

/    |  \  \  \  at  the  vertex  is  written  first,  followed  by 

/ J    i  \    \    \      a  short  dash,  and  then  the  letters  of  the 

// — f"  \     \ — /"  base.     Thus,  the  pyramid  shown  in  Fig. 

//  \      \  /     69  would  be  referred  to  as  the  pyramid 

1/  J    \       r        s-abed. 

//  \    /  Note  that  the  pyramid  in  Fig.  69  has 

L V  a  square  base;  the  sides  asb,  bsc,  csd,  and 

dsa  are  triangles  whose  vertexes  have  the 

Fig.  69.  . ° 

common  point  s.  The  polyedron  (pyra- 
mid) in  this  case  is  formed  by  the  intersection  of  five  planes — 
four  forming  the  sides  and  the  fifth  forming  the  base. 

If,  from  the  vertex  s,  a  perpendicular  be  drawn  to  the  base,  the 
line  so  in  Fig.  69,  this  line  is  the  altitude  of  the  pyramid. 


§2 


MENSURATION  OF  SOLIDS 


105 


If  the  base  of  a  pyramid  is  a  regular  polygon  and  the  projection 
of  the  vertex  upon  the  base  coincides  with  the  geometrical  center 
of  the  base,  the  pyramid  is  called  a  regular  pyramid.  Suppose 
that  the  base  of  the  pyramid  shown  in  Fig.  69  is  a  square  and 
that  the  point  o,  which  is  the  projection  of  the  vertex  s  upon  the 
base,  is  the  center  of  the  square,  then  s-abed  is  a  regular  pyramid, 
and  the  line  so  is  called  the  axis  of  the  pyramid.  The  length  of 
so  is  also  equal  to  altitude  of  the  pyramid. 

If  a  line  sf  be  drawn  from  the  vertex  s  perpendicular  to  one  of 
the  sides  of  the  base  of  a  regular  pyramid,  this  line  is  called  the 
slant  height.  There  will  be  as  many  slant  heights  as  there  are 
sides  in  the  polygon  forming  the  base,  and  for  regular  pyramids, 
all  slant  heights  are  equal ;  if  the  pyramid  is  not  regular,  some  or  all 
the  slant  heights  are  different. 

114.  Area  and  Volume  of  Pyramid. — The  lateral  area  of  any 
regular  pyramid  is  equal  to  perimeter  of  the  base  multiplied  by 
one-half  the  slant  height.     This  is  evident,  since  the  sides   are 

isosceles  triangles,  and  the  area  of  one  side  is    ~  X  be,  Fig.  69; 
hence,  the  lateral  area  is  ~-  X  be  X  n  (n  =  number  of  sides  in 


the  base).     But  n  X  be  —  p,  the  perimeter  of  the  base. 

s  =  the  slant  height  and  A  =  the  lateral  area; 

then, 


Let 


A  =2Xp 


hp 


If  the  pyramid  is  not  regular,  calculate  the 
area  of  each  face  and  find  the  sum. 

115.  The  volume  of  any  pyramid  is  equal  to 
the  area  of  the  base  multiplied  by  one-third  the 
altitude.  Let  h  =  the  altitude  =  so  in  Figs. 
69  and  70,  V  =  the  volume,  and  a  =  the  area 
of  the  base;  then, 

h 
3 

Example. — The  sides  of  the  base  of  a  regular  triangular  pyramid  are 
13J4  in.  long  and  the  altitude  is  18%  in.;  what  is  the  entire  area  and  what  is 
the  cubical  contents  of  the  pyramid. 

Solution. — Fig.  70  shows  a  sketch  of  the  pyramid.  Here  sf  is  the  slant 
height  and  so  is  the  altitude.  The  base  abc  is  an  equilateral  triangle,  and 
of  is  the  apothem,  the  length  of  which  is  (see  table,  Art.  81)  .28868  X  13.25 
=  3.825  in.     The  triangle  sof  is  a  right  triangle,  right-angled  at  o,  and  sf 


V 


X  a  =  \ha 


106        ELEMENTARY  APPLIED  MATHEMATICS  §2 

=  V3^825*  +  18.3752  =  18.769   in.     The   lateral   area   is,    by  formula  of 

Art.  114,  A  =   18'^69  X  13.25  X  3  =  373.034  sq.  in.     The  area  of  the  base 

13  2"> 
is,  by  formula  (2)  of  Art.  69,     ■  7— V3  =  76.021  sq.  in.     The  entire  area 

therefore  is  373.03  +  76.02  =  449.05  sq.  in.     Ans. 
The  volume  is  given  by  the  formula  above,  and  is 

V  =       ,f       X  76.02  =  465.62  cu.  in.     Ans. 
o 

116.  Frustum  of  a  Pyramid. — If  a  pyramid  be  intersected  by 
a  plane  parallel  to  the  plane  of  the  base  and  the  top  part  removed, 
the  remaining  portion  of  the  pyramid  is  called  a  frustum  of  the 
pyramid.     Thus,   referring   to   Fig.    71,   the   plane   a'b'c'd'e'   is 

.  parallel  to  the  base  abcde;  now  removing  that 

part  of  the  pyramid  above  the  intersecting 
It  i\\\  plane  (here  indicated  bv  the  dotted  lines),  the 

i-4-m  remaining  part  is  a  frustum  of  the  pyramid 

■j   \  yf  s— abcde.     The  two  bases  of  the  frustum  are 

l__JpA        similar  plane  figures  (see  Art.  139),  and  they 
i     \    x\      must  be  parallel. 
6      \  /  A  good  example  of  a  frustum  of  a  pyramid 

y  is  the  bottom  of  a  bin;  also,  a  spout.     Here  the 

frustum  is  inverted,  the  large  end  being  up, 
and  the  ends  are  open.      In  such  cases,  the 
frustum  has  the  shape  of  a  solid  that  would  exactly  fit  the 
opening. 

The  altitude  of  a  frustum  is  the  perpendicular  distance  between 
the  bases;  it  is  indicated  in  Fig.  71  by  the  dotted  line  o'o,  which 
is  a  part  of  the  altitude  of  the  pyramid. 

If  the  frustum  is  a  part  of  a  regular  pyramid,  the  slant  height  of 
the  frustum  is  that  part  of  the  slant  height  of  the  pyramid  that  is 
included  between  the  bases. 

117.  In  problems  relating  to  frustums,  the  altitude  and  the 
Length  of  the  sides  of  both  bases  are  usually  given.  For  a  regular 
pyramid,  the  sides  are  equal  trapezoids,  and  the  lateral  area  is  n 
times  the  area  of  one  side,  where  n  =  the  number  of  sides.  Let 
p'  =  the  perimeter  of  the  lower  (larger)  base  and  p"  =  the 
perimeter  of  the  upper  base;  then,  if  s  =  the  slant  height  of  the 
frustum  of  a  regular  pyramid  and  A  =  its  lateral  area, 

.-1  =  i«(p'  +  p") 

118.  To  find  the  volume  of  any  frustum  of  a  pyramid,  whether 
a  regular  pyramid  or  otherwise,  let  a'  =  area  of  lower  base, 


§2  MENSURATION  OF  SOLIDS  107 

a"  =  area  of  upper  base,  h  =  the  altitude,  and  V  =  the  volume 
of  the  frustum;  then, 

V  =  lh(a'  +  a"  +  vV  X  a") 
That  is,  the  volume  of  a  frustum  of  any  pyramid  is  equal  to  one-third 
the  altitude  multiplied  by  the  sum  of  the  areas  of  the  two  bases  and  the 
square  root  of  their  product. 

Example. — The  lengths  of  the  edges  of  the  lower  base  of  a  frustum 
of  a  triangular  pyramid  (one  having  three  sides)  are  12  in.,  15  in.,  and  18  in.; 
the  corresponding  edges  of  the  upper  base  are  7  in.,  8%  in.,  and  10J^  in.; 
if  the  height  of  the  frustum  is  10  inches,  what  is  its  volume? 

Solution. — Using  the  formula  of  Art.  70,  the  area  of  the  upper  base  is 
A  =  J  \/26.25(26.25  -  2  X  7)  (26.25  -  2  X  8.75)  (26.25  -  2  X  10.5) 
=  30.385  sq.  in.  since  p  =  7 '  +  8.75  +  10.5  =  26.25  in. 

Using  the  same  formula,  area  of  lower  base  is  89.294  sq.  in. 

Then,  volume  of  frustum  is 
V  =  §  X  10(89.29-1  +  30.385  +  V89.294  X  30.385)  =  572.56  cu.  in.  Ans. 

119.  Prismatoids. — A  prismatoid  is  a  polyedron  whose  ends  are 
parallel,  but  the  polygons  forming  the  outline  of  the  ends  are  not 
equal  and  may  have  a  different  number  of 
sides.  In  Fig.  72  is  shown  a  prismatoid, 
one  end  of  which  is  a  pentagon  and  the 
other  end  a  quadrilateral.  It  will  be  noted 
that  four  of  the  sides,  abgf,  aeif,  eihd,  and 
cdhg  are  quadrilaterals,  while  the  side  bgc  is 
a  triangle.  These  sides  are  all  plane  sur- 
faces, and  are  formed  by  passing  planes 
through  the  edges  of  the  lower  base  and  the 
vertexes  of  the  upper  base.  Thus,  planes 
passed  through  the  vertex  g  and  the  edges 
be  and  cd  intersect  to  form  the  lateral  edge 
gc;  the  intersection  of  these  two  planes  with 
planes  passing  through  the  edges  ab  and  de  Flo  72. 

and  the  vertexes  g  and  /  and  h  and  i  inter- 
sect in  the  lateral  edges  gb  and  hd;  a  fifth  plane  passed  through 
ae  and  the  vertexes  /  and  i  intersects  the  last  two  planes  in  the 
lateral  edges  fa  and  ie. 

If  both  ends  have  the  same  number  of  sides  and  are  similarly 
situated,  the  prismatoid  is  called  a  prismoid.  Thus,  the  frustum 
of  a  pyramid  is  a  prismoid,  see  Fig.  71,  since  the  ends  are  parallel 
and  both  contain  the  same  number  of  sides.  Fig.  73  shows 
another  prismatoid,  the  sides  ade  and  bef  being  triangles;  one 
end,  abed,  is  a  quadrilateral  and  the  other  end  is  a  right  line  ef. 


108         ELEMENTARY  APPLIED  MATHEMATICS 


§2 


To  find  the  volume  of  a  prismatoid,  let  h  =  the  altitude  =  the 
perpendicular  distance  between  the  parallel  ends,  let  a'  =  area 
of  lower  base,  a"  =  area  of  upper  base,  and  am  =  area  of  the 
middle  cross  section;  then  the  volume  of  the  prismatoid  is  equal  to 
one-sixth  the  altitude  multiplied  by  the  sum  of  the  areas  of  the  upper 
and  lower  bases  and  4  times  the  area  of  the 
middle  cross  section;  that  is, 


V 


=  £(<*'  +  a' 

b 


+  4am) 


To  find  the  area  of  the  middle  section, 
note  that  a',  Fig.  72,  is  midway  between  a 
and  /,  b'  is  midway  between  b  and  g,  etc. ; 
fg  +  ab  h,  ,  =  bc+  0  _  6c 
!        =  2' 


hence,  a'b'  = 


b'c'  = 


Fig 


c'd' 


gh  +  cd 


j  etc.     Knowing  the  lengths 


of  the  sides  and  their  position  and  directions,  the  area  can  be 

found. 

This  formula  is  called  the  prismoidal  formula;  it  may  be  used 

to  calculate  the  volumes  of  many  solids  besides  prismatoids. 

Applying  it  to  the  example  of  the  last  article,  the  sides  of  the 

-am          f               7+12       ng    15  +  8.75 
middle    section    are   — ^ —   =  "-5, ~ —       =    11.875,    and 

— n —       =  14.25;  the  area  of  this  triangle  is  55.964  sq.  in. 

Therefore,  the  area  of  the  prismoid 
(frustum)  is  V  =  %X  10  (89.294  +  30.385 
+  4  X  55.964)  =  572.56  cu.  in.. 

120.  The  Wedge.— When  the  base  of  the 
prismatoid  is  a  rectangle  and  the  end 
parallel  to  it  is  a  right  line  parallel  to  one 
of  the  edges  of  the  base,  the  prismatoid  is 
called  a  wedge.  Referring  to  Figs.  73  and 
74,  if  the  base  abed  is  a  rectangle  and  ef 
is  parallel  to  ab,  the  figure  represents  a 
wedge.  In  Fig.  73,  the  upper  base,  the 
line  ef,  is  shorter  than  ab  =  dc,  and  the  solid  is  a  prismatoid; 
in  Fig.  74,  ef  =  ab  =  dc,  and  the  solid  is  a  triangular  prism 
whose  liases,  eda  =  feb,  are  parallel,  and  whose  altitude  is  ab. 

The  volume  of  the  prismatoid  in  Fig.  73  may  be  found  by 
applying  the  prismoidal  formula;  but  an  easier  method  is  to  find 


Fig.  74. 


§2  MENSURATION  OF  SOLIDS  109 

the  sum  of  parallel  sides,  divide  it  by  3,  and  multiply  the  quotient 
by  the  area  of  a  right  section —  one  taken  at  right  angles  to  the 
parallel  sides.  Thus,  letting  a  =  the  area  of  the  right  section 
a'e'd', 

y  _  a  X  (ab  +  dc  +  ef)  ,  , 

o 

When  the  prismatoid  becomes  a  wedge,  ab  =  dc,  and 
y  =  a  X  (2ab  +  ej) 

o 
When  a  wedge  becomes  a  prism,  ab  —  dc  =  ef,  and 

V  =  a  X  ab  =  ah 
The  volume  of  the  wedge  may  also  be  calculated  by  the  prismoi- 
dal  formula.     Thus,  let  ab  =  m,  be  =  n,  ef  =  m',  and  h  =  the 
altitude;  then,  for  the  middle  section,  the  sides  parallel  to  ab  or  dc 

are  equal  to  — = —  5  the  sides  parallel  to  be  or  ad  are  equal  to 

— n —  =  «;  4  times  the  area  of  the  middle  section  is „ — 

X  n  X  4  =  mn  +  m'n;  the  area  of  one  end  is  mn,  and  the  area  of 
the  other  end  is  the  line  ef  =  0;  therefore,  by  the  prismoidal 
formula,  V  =  A  mn  +  0  +  (mn  +  m'n)   or 

V  =  lh(2m  +  m')n.  (3) 

For  the  wedge  shown  in  Fig.  74,  m  =  m',  and 

V  =  \hmn.  (4) 


EXAMPLES 


(1)  How  many  cords  of  wood  are  contained  in  three  piles  having  the  fol- 
lowing dimensions:  30'  6"  X  7'  X  4',  20'  X  8'  X  4',  and  8'  X  5'  X  4? 
One  cord  contains  128  cu.  ft.  Arts.   12ff,  say  13  cords. 

(2)  A  spout  has  the  form  of  a  frustum  of  a  square  pyramid;  the  bases  are 
18  in.  and  10  in.  square,  and  the  altitude  is  28  in.  How  many  square  feet 
of  sheet  iron  will  be  required  to  make  this  spout?.  A  rw.    1 1  sq.  ft. 

(3)  How  many  gallons  will  the  spout  mentioned  in  the  last  example  hold 
when  filled?  A ns.  24.404  gal. 

(4)  A  trench  20  ft.  wide  and  36  ft.  long  is  dug;  the  bottom  of  the  trench 
is  level,  but  the  top  slopes,  me  end  of  the  trench  being  8  ft.  9  in.  deep,  the 
other  end  5  ft.  3  in.  deep,  and  the  slope  gradual  from  end  to  end.  If  the 
walls  and  ends  are  vertical,  how  many  cubic  yards  of  material  were  removed? 

Ans.  186j  cu.  yd. 


110        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


(5)  A  freight  car  is  loaded  with  wood  in  the  following  manner,  all  sticks 
being  cut  to  4-foot  lengths:  two  piles  at  each  end  of  the  car  placed  crosswise 
and  one  pile  in  the  middle  (divided  into  two  equal  parts)  placed  lengthwise 
of  the  car.  The  average  height  of  the  piles  at  one  end  is  7'  8",  at  the  other 
end  7'  6",  and  the  height  of  the  pile  in  the  middle  is  7';  the  inside  dimen- 
sions of  the  car:  length,  36  ft.;  width,  8  ft.  6  in.;  height  8  ft.  If  an  allow- 
ance of  18  in.  of  the  car  length  be  made  for  spacing  between  the  piles,  how 
many  cords  were  placed  in  the  car?  Ans.  16.3  cords. 

Suggestion — The  total  length  of  the  piles  is  36  —.1.5  =  34.5  ft.  The  length  of  each  pile 
placed  crosswise  'measured  lengthwise  of  the  car)  is  (34.5  —4)  -5-  2  =  15>£  ft.;  since  the 
length  of  the  middle  pile  is  4  ft. 

(6)  A  pedestal  has  a  lower  base  28"  X  40",  and  upper  base  21"  X  30", 
and  the  perpendicular  distance  between  the  bases  (which  are  parallel)  is 
28  in.     How  many  cubic  feet  are  in  the  pedestal?        Ans.  14.99.  say  15  cu.  ft. 

(7)  How  many  cubic  yards  of  concrete  are  required  to  make  a  foundation 
wall,  the  outside  dimensions  of  which  are:  length,  136  ft.;  breadth,  68  ft. 
6  in.,  the  height  of  the  wall  being  uniformly  7  ft.  9  in.  and  the  thickness 
being  18  in.?  Atis.  173.514  cu.  yd. 


THE  THREE  ROUND  BODIES 

121.  The  Cylinder. — If  one  side  ad  of  a  rectangle  abed, 
Fig.  75.  be  fixed  in  position  and  the  rectangle  be  revolved  around 

this  fixed  side,  the  opposite  side  cd  will 
generate  a  curved  surface  that  is  called  a 
cylinder.  The  other  two  sides  of  the  rec- 
tangle will  generate  the  circles  bb'b"b'"  and 
cc'c"c'",  which  form  the  ends  or  bases,  of 
the  cylinder.  It  will  be  seen  that  the  bases 
are  parallel  and  that  they  are  perpendic- 
ular to  the  line  ad  passing  through  their 
centers;  the  line  ad  is  called  the  axis  of  the 
cylinder.  Any  line  drawn  on  the  cylinder 
parallel  to  the  axis  is  called  an  element 
of  the  cylindrical  surface,  or,  simply,  an 
element;  in  Fig.  75,  be,  b'c',  etc.  are  ele- 
ments of  the  cylinder,  and  they  are  evi- 
dently positions  occupied  by  the  line  be  as 
it  revolves  about  the  axis  ab  while  generating  the  cylinder. 

122.  A  cylinder  generated  as  just  described  is  called  a  cylinder 
of  revolution,  and  since  the  bases  are  circles  and  perpendicular 
to  the  axis,  it  is  also  called  a  right  cylinder  with  circular  base. 

The  same  rules  and  formulas  that  were  given  for  finding  the 
area  and  volume  of  a  prism  may  be  used  to  find  the  area  and  vol- 


Fig.  7c 


§2 


MENSURATION  OF  SOLIDS 


111 


ume  of  a  cylinder,  but  since  all  right  sections  are  circles,  these 
rules  and  formulas  may  be  somewhat  simplified.  Referring  to 
Fig.  76,  suppose  the  cylinder  is  a  cylinder  of  revolution  and  that 
it  lies  on  a  plane  surface;  the  line  (element)  ab  will  then  be  the  line 
of  contact  of  plane  and  the  cylinder,  and  the  plane  is  said  to  be 
tangent  to  the  cylinder.  Now  roll  the  cylinder  on  the  plane  surface 
until  the  element  ab  again  comes  in  contact  with  the  plane;  the 
distance  moved  through  by  the  cylinder  will  evidently  be  equal 
to  the  circumference  of  a  circle  whose  diameter  is  equal  to  the 
diameter  of  the  base.     During  this  movement,  every  element  of 


rr(£- 


b^ 


Fig.  76. 

the  cylinder  has  come  into  contact  with  the  plane,  and  the  surface 
thus  touched  by  the  cylinder,  which  is  equal  to  the  outside  surface 
of  the  cylinder,  is  the  rectangle  aa'b'b,  called  the  development  of 
the  cylinder.  The  outside  area  of  a  cylinder,  corresponding  to 
the  lateral  area  in  a  prism,  is  called  the  convex  area,  and  the  sur- 
face is  called  the  convex  surface.  The  diameter  of  a  cylinder  is 
the  diameter  of  a  right  section,  and  the  altitude  of  a  cylinder 
is  the  perpendicular  distance  between  the  bases;  it  is  equal  to  the 
length  of  the  axis  in  a  cylinder  of  revolution,  and  it  is  represented 
by  h  in  Fig.  77.  Letting  d  =  the  diameter  of  the  cylinder,  I  =  the 
length  of  the  axis,  it  is  plain  from  Fig.  76  that  the  convex  area  of  a 
cylinder  of  revolution  is 

A  =  irdl 
If  the  cylinder  is  not  a  right  cylinder,  but  has  parallel  bases,  as 
in  Fig.  77,  it  may  be  made  a  right  cylinder  by  passing  a  plane 


112        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


through  the  cylinder  perpendicular  to  the  axis  and  just  touching 
one  edge  of  the  base  at  c;  the  part  ebjc  thus  cut  off  may  be  placed 
on  the  other  end,  as  indicated  by  the  dotted  lines,  and  the  cylinder 
then  becomes  a  right  cylinder  e'ecd.  The  length  of  the  axis  is 
the  Bame  aa  before,  that  is,  o'o  =  o'"o";  hence,  A  =  rdl,  as 
before. 


Fig. 


123.  If  one  of  the  bases  is  perpendicular  to  the  axis  and  the 
other  is  not,  for  instance,  if  the  cylinder  has  the  form  e'bcd, 
Fig.  77,  the  area  is  equal  to  the  sum  of  the  areas  of  the  right  cylin- 
der e'ecd  and  the  wedge-shaped  solid  cbe.  The  latter  is  evidently 
equal  to  one-half  of  a  right  cylinder  whose  base  is  the  circle 
ec  (center  e")  and  whose  axis  is  equal  to  the  element  eb;  that  is, 

the    area    of    ebc  =  ^  X  nd  X  eb  =  ird  X  oo",    since    %  X  eb 

e'h  _i_  cic 

TJni  II      •  II     III  III  c   u       \ 


=   00 


oo"  +  o"o' 


=    00 


hence,    area  of 


cylinder  e'bcd  is  A  =  ird  X  o"o'"  +  rd  X  oo"  =  ird(oo"  +  o"o'"), 
or 

fe'b  +  (Jc\ 


,  (e'b  +  rlc\ 


If  both  bases  are  oblique  to  the  axis,  cut  the  cylinder  by  planes 
perpendicular  to  the  axis,  calculate  the  areas  of  both  wedge- 
shaped  solids  and  of  the  right  cylinder  included  between  them, 
and  then  find  the  sum  of  the  three  results. 


§2 


MENSURATION  OF  SOLIDS 


113 


Example. — The  cylindrical  part  of  a  plater  roll  forms  a  right  cylinder 
17  in.  in  diameter  and  30  in.  long;  what  is  its  convex  area? 

Solution. — Using  the  formula  of  Art.  122, 

A  =  ir  X  17  X  36  =  1922  66  sq.  in.  =  13.352  sq.  ft.    Ans. 

124.  If  a  right  cylinder  be  intersected  by  a  plane  oblique  to 
the  axis,  which  cuts  the  base  of  the  cylinder,  the  part  thus  cut 
off,  ifgk,  Fig.  78,  is  called  a  cylindrical  ungula,  the  word  ungula 
having  reference  to  an  object  shaped 
like  a  horse's  hoof.  Let  fg  be  the 
line  in  which  the  cutting  plane  inter- 
sects the  base.  Bisect  fg,  and  draw 
k'k  through  the  point  of  bisection  o', 
and  perpendicular  to  fg;  then  k'k  is 
a  diameter  of  the  base.  Let  o  be 
the  middle  point  of  k'k,  and  ok  =  ok'9 
=  r,  the  radius  of  the  base  =  radius  of 
cylinder.  Let  o'f  =  o'g  =  a,  o'k  —  b, 
and  arc  gk  =  1/2  arc  fkg  =  </>;  then, 
the  convex  area  of  the  ungula  is,  when 
h  =  the  altitude  ik, 

A  =  **[<*  -  ,)  *  +  a]        (1) 

If  the  line  of  intersection  pass 
through  the  center  of  the  base,  as 
indicated  by  the  dotted  outline  in  Fig.  78,  b  and  a  are  both  equal 

to  r,  and  formula  (1)  then  becomes  A  =  —    [r  —  r)-  +  r\t 

from  which 

A  =  2rh  (2) 

If  the  cutting  plane  just  touches  the  other  edge  of  the  base,  as 
in  the  case  of  ebc,  Fig.  77,  b  =  2r  and  a  =  0.     Substituting  these 

values  in  formula  (1),  A  =  -^7   (2r  —  r)^  +  0|,  since  0  is  then 
a  semicircle  and  is  equal  to  7rr.     Reducing  this  expression, 

A  =  vrh  =  Vz-wdh  (3) 

which  is  the  same  value  as  was  obtained  in  Art.  123.     Note  that 

-  is  the  angle  okg  in  radians. 

Example. — Suppose  the  radius  of  a  right  cylinder  is  9  in.  and  that  tin- 
cylinder  is  cut  by  a  plane  in  such  manner  that  the  altitude  of  the  ungula  is 
8  in.;  if  a  =  6>£  in.,  what  is  the  convex  area  of  the  ungula? 


114        ELEMENTARY  APPLIED  MATHEMATICS  §2 

Solution. — Referring  to  Fig.  78,  2a  =  2  X  6.5  =  13  =  fg  =  chord  of  arc 
fkg;  b  =  ok  =  height  of  arc  =  h  =  r  —  >2\/4r2  —  c-  =  2.775,  by  formula 

d>  I 

(3),  Art.  87;  -  =  one-half  the  length  of  the  arc  divided  by  the  radius  =  =-■ 

Since  t  =      =     ,.,      =  .21346,  which  is  very  near  .21,  use  formula  (3)  of 

Art.  94,  and      =      -  =  „  ,,  .    ,   OQj..     =  .80/04-.     .Nuw      substituting      in 
r        2r       2r(15  4-  obt2) 

fcinnula  (1), 

A  =  2   9  ry?  8K2-775  _  9)0-807  +  6.5]  =  76.61  sq.  in.     An*. 

125.  The  volume  of  any  cylinder  whose  bases  are  parallel  is 
equal  to  the  area  of  the  base  multiplied  by  the  altitude;  it  is  also 
equal  to  the  area  of  a  right  section  multiplied  by  the  length  of 
the  axis.  If  the  cylinder  is  one  of  revolution,  the  base  is  equal 
to  a  right  section,  and  both  are  equal  to  the  area  of  a  circle  whose 
diameter  is  equal  to  the  diameter  of  the  cylinder;  likewise,  the 
altitude  is  equal  to  the  length  of  the  axis.  Let  h  =  the  altitude, 
/  =  the  length  of  the  axis,  d  =  the  diameter  of  the  cylinder, 
a  =  the  area  of  the  base;  then, 

y  =  ajl=*d2l  (!) 

4 

For  a  cylinder  of  revolution, 

V  =  nrVi  =  -.  d-h  (2) 

4 

When  the  bases  are  not  perpendicular  to  the  axis,  they  have 
the  form  of  an  ellipse,  and  their  areas  may  be  calculated  by  formula 
(1),  Art.  99.  Thus,  referring  to  Fig.  77,  the  lower  base  bjck  is 
an  ellipse,  whose  major  axis  (long  diameter)  is  be,  and  whose  minor 
axis  (short  diameter)  is  jk.  Knowing  these  two  dimensions 
(which  call  D  and  d,  respectively),  a  in  formula  (1),  Art.  99,  is 

2  and  b  is  ~;  hence  the  area  of  the  ellipse  (base)  is  expressed  by 
irab  =  7T  X  2"  X  2=  i^d,  and  the  volume  of  the  cylinder  is 

T  =  *Ddh  (3) 

4 

126.  The  volume  of  an  ungula  is  given  by  the  following 
formula,  in  which  the  letters  have  the  same  meaning  as  in  formula 
(1),    Art.    124: 


§2 


MENSURATION  OF  SOLIDS 


115 


If  the  line  of  intersection  fg  of  the  cutting  plane  with  the  plane 
of  the  base  passes  through  the  center,  as  indicated  by  the  dotted 
outline  i'k'g'f,  Fig.  78,  a  =  b  =  r,  and  formula  (1)  reduces  to 

V  =  §  r*h  (2) 

If  the  cutting  plane  just  touches  the  edge  of  the  base,  as  indi- 
cated by  ebjck,  Fig.  77,  a  =  0,  b  =  2r,  </>  =  irr,  and  formula 
(1)  reduces  to 

(3) 


V  =  \r--h 


That  forniula  (3)  is  correct  is  readily  seen,  since  the  ungula  ebc 
is  equal  to  one-half  the  cylinder  whose  base  is  the  circle  ec  and 
whose  altitude  is  eb  =  h  in  formula  (1). 

Example  1.— The  shaft  of  a  plater  roll  is  65  in.  long  and  8  in.  in  diameter; 
if  made  of  steel,  one  cubic  inch  of  which  weighs  .2836  lb.,  what  is  the  weight 
of  the  shaft? 

Solution. — The  shaft  is  a  cylinder  of  revolution,    and   its   volume   is 

F=:(J!I=7X82X65  =  .7854   X  64    X  65  =  3267.264  cu.  in.     The 

4  4 

weight  of  the  shaft  is  3267.264  X  .2836  -  926.6-  lb.     Ans. 

Example  2. — Referring  to  the  example  in  Art.  124,  what  is  the  volume  of 
the  ungula? 

Solution.— Here    r  =  9,    h  =  8,    a  =  6.5,    b    (found    by    calculation) 

=  2.775,  and  -  (found   by    calculation)  =  .807,  or  <t>   =  .807  X  9  =  7.263. 
r 

Substituting  these  values  in  formula  (1 ), 

V  =    3  x  28775  [6.5  (3  X  92  -  6.52)  +  3 
X9(2.775 -9)7.263]  =80.86  cu.  in.     Ans. 

127.  If  the  cylinder  has  a  hole 
through  it,  it  is  called  a  hollow 
cylinder,  pipe  or  tube.  A  cross 
section  of  a  hollow  cylinder  is 
shown  in  Fig.  79.  It  is  assumed 
that  the  hole  is  also  a  cylinder  and 
that  its  axis  coincides  with  the  axis 
of  the  cylinder  that  contains  the 
hole.     The  volume  of  the  hollow 

cylinder  is  evidently  equal  to  the  difference  of  the  volumes  of  the 
cylinder  and  hole.  If  I  =  the  length  of  the  cylinder,  R  =  radius 
of  cylinder,  r  =  radius  of  hole, 

V  =  wRH  -  irrH  =  tI(R2  -  r2)  =  irl(R  +  r){R  -  r) 


Pio.  79. 


«|Z(Z> +  d)(0 


d) 


(1) 


116        ELEMENTARY  APPLIED  MATHEMATICS  §2 

Suppose  the  cross  section  in  Fig.  79  is  that  of  a  tube  of  thickness 
/;  then  t  =  R  —  r  =  oa :  —  ob.  Suppose  further  that  a  circle 
be  drawn  midway  between  the  inner  and  outer  circles  as  indicated 
by  the  dotted  line.     Then  the  radius  oc  of  this  circle  is  equal  to 

rm  =   — s —  J    its  length  (circumference)   is  2irrm  =  2tt  X  — « — 

=  r{R  +  r);  multiplying  this  by  the  thickness  t  and  length  I, 
the  product  will  be  the  volume  of  a  flat  plate  having  the  same 
cubical  contents  as  the  tube.     Or,  since  t  =  R  —  r, 

V  =  2irrm  tl  =  t(R  +  r)  (R  -  r) 
which  is  the  same  as  formula  (1).     Letting  2rm  =  dm  =  diameter 
of  middle  circle, 

V  =  irdmtl  (2) 

In  other  words,  the  cubical  contents  of  a  hollow  cylinder,  pipe, 
or  tube  is  equal  to  the  continued  product  of  r  =  3 .  1416,  the  mean 
diameter,  the  thickness,  and  the  length  of  the  cylinder. 

Example  1. — A  cylindrical  tank  made  of  wrought  iron,  a  cubic  inch  of 
which  weighs  .2778  lb.,  is  to  hold  1000  gallons;  the  tank  is  to  stand  with 
the  axis  vertical,  and  the  diameter  (inside)  is  to  be  the  same  as  the  height; 
what  is  the  diameter  of  the  tank?  If  the  thickness  of  the  shell  is  J^  m-> 
what  is  its  weight? 

Solution. — Since  the  diameter  equals  the  height,  d  =  I  in  formula  (1), 
Art.  125;  hence,  since  there  are  231  cu.  in.  in  a  gallon, 

t,         }iwd2d        ird3        1fiAn   „     ,  a/924000         aa  cno  „,,,.         t 

V  =  -2gy-  =  g24  =  1000,  or  d  =  ^|3  U1g    =  66.503,  say  66}£  in.    Am. 

The  inside  diameter  is  66.5  in.,  and  since  the  thickness  is  .5  in.,  the  out- 
side diameter  is  66.5  +  2  X  .5  =  67.5  in.;  then,  by  formula  (1),  Art.  127, 

the  cubical  contents  of  the  shell  is  V  =  7  X  66.5(67.5  +  66.5)  X  1  =  6998.7 

cu.  in.;  this  multiplied  by  .2778.  the  weight  of  a  cubic  inch  of  wrought  iron, 
is  6998.7  X  .2778  -  1944.24.  say  1944  lb.     Ans. 

Example  2. — Assuming  that  the  shell  in  the  last  example  is  made  by  roll- 
ing a  flat  sheet,  %  in-  thick,  into  a  cylindrical  form,  what  should  be  the 
size  of  the  sheet  ? 

Solution. — The  shape  of  the  sheet  will  be  that  of  a  rectangle,  one  side  of 
which  is  equal  to  the  height  of  the  tank  =  66.5  in.,  and  the  other  side  will 
be  equal  to  the  circumference  of  a  circle  that  is  midway  between  the  inside 
and  outside  circles  of  a  right  section  of  the  tank.  The  diameter  of  this 
circle  is  evidently  equal  to  the  inside  diameter  plus  the  thickness,  or  66.5 
+  .5  =  67  in.,  and  its  circumference  is  3.1416  X  67  =  201.49,  say  201^  in. 
Therefore,  the  sheet  must  be  201 H  m-  long  and  66>2  in-  wide.     Ans. 

N"te  that  the  mean  diameter  of  the  two  circles  is  — '- — —  =  67  in., 

the  same  result  as  was  obtained  by  adding  the  thickness  to  the  inside 
diameter. 


§2 


MENSURATION  OF  SOLIDS 


117 


128.  The  term  cylinder  is  not  confined  to  solids  having  cir- 
cular bases  or  whose  right  sections  are  circles;  it  is  applied  to 
solids  whose  bases  have  any  shape  whatever,  except  those  classed 
as  prisms,  the  elements  of  whose  surfaces  are  all  perpendicular 
to  a  right  section.  Such  solids  may  all  be  generated  by  keeping 
a  line  in  contact  with  the  outline  of  a  plane  figure,  then  moving 
the  line  so  that  it  touches  every  point  of  the  plane  figure,  and  at 
the  same  time,  always  remains 
parallel  to  a  given  right  line. 
The  outline  of  the  plane  figure 
is  called  the  directrix,  and  the 
moving  line  is  called  the  gen- 
eratrix. In  Fig.  80,  abedefgh 
is  the  directrix  and  a' a  is  the 
generatrix.  The  generatrix 
moves  over  the  directrix, 
always  remaining  parallel  to 
the  fixed  line  AB,  and  thus 
generates  the  cylindrical  sur- 
face shown  in  the  figure. 
While,  strictly  speaking,  the/ 
directrix  should  always  be  a 
curve  in  order  to  apply  the 
term  cylinder  to  the  surface 

generated,  it  may,  nevertheless,  be  applied  to  those  surfaces  in 
which  the  generatrix  consists  of  both  straight  lines  and  curves. 

Keeping  in  mind  this  definition  of  a  cylindrical  surface  and  call- 
ing the  solid  that  it  bounds  a  cylinder,  the  general  formulas 
previously  given  for  the  area  and  volume  of  a  cylinder  will  apply 
in  this  case  also. 

Example.— What  volume  of  stock  is  displaced  by  a  beater  roll  48  in. 
wide  and  60  in.  in  diameter  that  is  immersed  to  a  depth  of  28  in.  in  the  stock  ? 

Solution.— The  outline  of  the  stock  displaced  is  a  cylinder  48  in.  long, 
with  equal  bases  having  the  shape  of  a  segment  of  a  circle;  and  a  right 
section  of  this  cylinder  is  a  segment  of  a  circle,  the  radius  of  the  circle  be.ng 
GO  -  2  =  30  in.,  and  the  height  of  the  segment  being  28  in.  '1  he  volume 
displaced  is  equal  to  the  area  of  the  segment  multiplied  by  the  length  of 

the  cylinder.  . 

To  find  the  area  of  the  segment,  first  calculate  the  chord,  using  formula 
(5),  Art.  87,  and  c  =  2V(2  X  30  -  28)28  =  59.867  in 

Since  the  angle  is  very  large,  use  formulas  (4)  and  (6)  of  Art.  94  to  find 
the  length  of  the  arc.     From  formula  (4),  t'  =  .19740+,  and  from  formula 


Fig.  80. 


118        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


90  25  X  30 
(5),  I  =  90.25  in.     The  area  of  the  sector  is      '     2 =  1353.75  sq.  in.; 

area  of  triangle  =—       .,      —  =    59.867   sq.    in.;    and   area  of  segment  = 

1353.75  -  59.87  =  1293.88  sq.  in.     Therefore,  volume  of  cylinder  =  volume 
of  stock  displaced  =  1293  88  X  48  =  62,106  cu.  in.     Ans. 

129.  Whenever  a  cylindrical  tank  is  partly  filled  with  a  fluid  of 
some  kind  and  is  so  placed  in  position  that  its  axis  is  neither 
vertical  nor  horizontal,  an  ungula  is  formed.  The  upper  surface 
of  the  fluid  is  always  a  horizontal  plane,  and  the  intersection  of 
this  plane  with  the  cylindrical  surface  forms  either  an  ungula  or 
what  might  be  called  a  frustum  of  an  ungula.  Thus,  referring  to 
Fig.  81,  which  shows  a  cylindrical  tank  having  one  end  higher 


Fig.  81. 

than  the  other  and  partly  filled  with  some  fluid  (stock,  for  ex- 
ample), the  upper  surface  aa'b'b  is  a  plane,  level  and  parallel  to 
the  horizontal  line  eg.  If  the  tank  were  extended  until  the 
plane  of  the  top  surface  just  touched  the  bottom  of  one  end,  as 
indicated  by  the  dotted  lines,  the  outline  of  the  fluid  contents 
would  be  the  ungula  f-acb.  The  part  that  is  actually  in  the  tank 
has,  in  this  case,  the  outline  acbb'c'a',  which  may  be  called  a 
frustum  of  the  ungula  f-acb,  and  its  volume  is  evidently  equal  to 
the  difference  between  the  volume  of  the  ungula  f-acb  and  the 
volume  of  the  ungula  f-a'c'b'.  The  point  /  can  be  easily  found 
when  the  length  of  the  tank  and  the  depth  of  the  fluid  at  each 
end  are  known.  Thus,  let  /  =  length  of  tank,  m  =  dc  =  depth 
at  lower  end,  n  =  d'e'  =  depth  at  upper  end,  and  x  =  c'f  =  the 
additional  length  of  tank  required;  then,  from  the  similar  triangles 
fed  and  fe'd',  cf  :  c'f  =  cd  :  c'd'.  But  c'f  =  x  and  cf  =  I  +  x; 
hence,  I  +  x:  x  =  m:  n,  or 

nl 

x  — ■ 

m  —  n 


§2  MENSURATION  OF  SOLIDS  119 

Example. — Referring  to  Fig.  81,  suppose  that  the  tank  is  42  in.  in  diam- 
eter, 12  ft.  6  in.  long,  and  that  the  upper  end  is  10  in.  higher  than  the  lower 
end;  if  partly  filled  with  stock,  so  that  the  distance  ed,  measured  along  the 
tank  bottom,  is  1434  m->  what  is  the  volume  of  stock  in  the  tank? 

Solution. — Before  the  point/  can  be  found,  it  is  necessary  to  calculate 
the  distance  c'd'  =  n  in  the  formula  above.  Since  c'c  =  12  ft.  6  in.  =  150  in. 
and  c'g  =  10  in.,  eg  =  Vl50*  -  102  =  149.67  in.  Draw  ck  perpendicular 
to  df;  then  dkc  is  a  right  triangle.  Since  dc  is  perpendicular  to  cf,  angle  dek 
=  angle  e'eg,  because  ck  is  perpendicular  to  eg,  which  is  parallel  to  df.  Con- 
sequently, triangles  dkc  and  c'gc  are  similar,  and  ck  :  cd  =  eg  :  c'c,  or  ck  :  42 

-  14.25  =  149.67  :  150,  from  which  ck  =  27.689  in.  =  distance  from  stock 
level  to  horizontal  line  eg.  Draw  c'k'  perpendicular  to  df;  then,  triangle 
d'k'c'  is  similar  to  triangle  dkc,  since  the  sides  are  parallel,  and  k'g  =  kc  = 
27.689  in.  But  k'c'  =  27.689  -  10  =  17.689  in.  Then,  from  the  similar 
triangles  d'k'c'  and  c'gc,  c'd'  :  150  =  17.689  :  149.67,  or  c'd'  =  17.728  in.  In 
the  formula  above,   I  =  150,   m  =  42  -  14.25  =  27.75,   and   n  =  17.728; 

150  X  17  728 
hence,   x  =  2775-17728  =  26534    in-  =  c'f>    and   tf  =  265-34  +  150  = 
415.34  in. 

Calculating  first  the  volume  of  the  ungula  f-acb,  it  is  evident  that  the  arc 
acb  is  greater  than  a  semicircle,  since  ed  is  less  than  the  radius  oe  =  42  -f-  2 
=  21  in.  Hence,  to  find  the  length  of  the  arc  acb,  find  the  length  of  the  arc 
aeb  and  subtract  it  from  the  circumference  of  which  it  is  a  part.  According 
to  Art.  87,  db2  =  ed  X  dc,  or  db  =  Vl4.25  X~27.75  =  19.886  in.  =  a  in 
formula  (1),  Art.  126.     Since  the  arc  is  very  large,  use  formula  (2)  of  Art. 

14  25 
94  to  find  the  length  of  the  arc  aeb.     Here  t  =  -  -,    '" '         =  .:;.">v_".t,  and 

.      40  X  21  X  .35829(15  +  16  X  .35S292)      co  0)0         ' 

l=     75  +  180  X  .35829*  +  64  X  .35829°     =52242     ,n'     fhc     c'rcumfer" 

ence  of  the  circle  is  42  X  3.1416  =  131.947  in.;  therefore,  arc  acb  =  131.947 

-  52.242  =  79.705  in.  =  2<t>  in  formula  (1),  Art.  126,  or  <t>  =  39.8525  in., 
say  39.853  in.  Using  this  formula,  h  =  415.34,  b  =  27.75,  a  =  19.886, 
r  =  21,  <fi  =  39.853,  and 

V  =    o^olX*  [19-886  (3  X  21 2  -  19.8862)  +  3  X  21(27.75  -  21)39.853 

=  176,577  cu.  in. 

Next  calculate  the  volume  of  the  ungula  f-a'c'b'.  The  arc  a'r'h'  is  less 
than  a  semicircle,   because    c'd'  is  less   than   the  radius  o'c'.     Here   d'b' 

1 7  70s  / 

=  Vl7.728(42  -  17.728)  =  20.744;  t  =   2  X20744    =  -42730'  and  *  =  2 
40  X  21  X  .4273(15  +  16  X  .42732)  =  2g      "  . 


2  X  75  +  180  X  .42732  +  64  X  .42736 
Now  using  formula  (1),  Art.  126,  h  =  265.34,  b  =  17.728,  a  =  20.744, 
r  =  21,  <f>  -  29.711,  and 

V"  =  Q    ™?%;1%^  [20.744(3    X    212    -  20.7442)   +  3  X  21(17.728  -  21; 
o    X   17 .7 Zo 

X  29.711]  =  61,831  cu.  in. 
Finally,  V  =  V  -  V"  =  176,577  -61831  =  114,746,  say  114,750  cu.  in., 
the  volume  of  the  stock  in  the  tank.     Ans. 


120        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


The  reader  will  find  it  excellent  practice  to  work  this  entire 
example,  performing  all  the  operations  herein  indicated. 

130.  The  Cone. — If  a  right  triangle  be  revolved  about  one  of 
its  legs  as  an  axis,  the  hypotenuse  will  generate  a  conical  surface, 
and  the  triangle  as  a  whole  will  generate 
a  solid  called  a  cone.  Thus,  referring  to 
Fig.  82  let  soa  be  a  right  triangle,  right- 
angled  at  o,  and  suppose  the  triangle  to 
be  revolved  about  the  leg  so;  then,  the 
hypotenuse  sa,  and  the  triangle  as  a 
whole  will  generate  the  solid  s-aa'a"b. 
The  point  s  is  called  the  vertex  of  the 
cone.  Any  right  line  drawn  from  s  to 
the  base  aa'a"b  is  called  an  element  of 
the  cone,  more  properly,  an  element  of 
the  conical  surface;  thus,  sa,  sa',  sa", 
etc.  are  elements  of  the  cone  in  Fig.  82. 
The  line  so  is  the  axis  of  the  cone,  and  o 
is  the  center  of  the  base  of  the  cone. 
Any  cone  generated  in  this  manner  is  a  cone  of  revolution;  and 
because  the  base  is  perpendicular  to  the  axis,  the  cone  is  also 
called  a  right  cone. 

Let  a  right  cone  be  laid  on  a  plane,  and  suppose  the  line  of  con- 
tact of  the  plane  and  cone  to  be  sa,  Fig.  82;  now  roll  the  cone  on 


Fig.  82. 


Fig.  83. 


the  plane,  the  vertex  s  remaining  stationary.  Each  element  of 
the  cone  will  come  into  contact  with  the  plane,  and  since  they  are 
all  of  the  same  length,  the  surface  of  contact  thus  generated  will 


§2  MENSURATION  OF  SOLIDS  121 

be  the  circular  sector  saa',  Fig.  83,  the  radius  sa  being  an  clement 
of  the  cone,  and  the  length  of  the  arc  aa'  will  be  equal  to  the  cir- 
cumference of  the  base  of  the  cone,  or  2ir  X  oa,  Fig.  82.  Since 
the  area  of  a  sector  is  equal  to  one-half  the  product  of  the  radius 
and  the  length  of  the  arc,  it  follows  that  the  convex  area  of  a  right 
cone  is  equal  to  one-half  the  product  of  the  perimeter  of  base  of 
the  cone  and  the  length  of  an  element.  The  length  of  an  element 
is  called  the  slant  height  of  the  cone;  representing  this  by  s,  and 
letting  r  =  the  radius  of  the  base,  the  convex  area  is 
A  =  }^s  X  2-n-r  =  irrs 
131.  If  a  cone  be  cut  by  a  plane  that  intersects  the  element 
directly  opposite  the  element  first  cut  by  the  plane,  the  surface 
formed  by  the  intersection  of  the  plane  and  cone  is  an  ellipse, 
except  when  the  plane  is  perpendicular  to  the  axis;  in  this  latter 
case,  the  section  is  a  circle,  which  is  an  ellipse  having  all  its  di- 
ameters equal.  Referring  to  Fig.  84,  the  section  a'c'b',  which  is 
formed  by  the  intersection  of  a  plane  perpendicular  to  the  axis  so 
of  the  cone,  is  a  circle;  the  base,  which  is  » 

also  perpendicular  to  the  axis,  is  also  a  /f\ 

circle.     In  Fig.  85,  the  intersecting  plane  /  j  \ 

is  not  perpendicular  to  the  axis  so  of  / 

the  cone,  but  it  intersects  the  element  /'.--|--.\ 

ab,  which  is  directly  (diametrically)  op-  jfvll  Z^K 

posite  the  element  dc;  hence,  the  section  /         W      \ 

aed  is  an  ellipse,  and  the  base  is  also  /  \ 

an  ellipse.  /  \ 

The  altitude  of  a  cone  is  the  perpen-       /-''""  ~^-\ 

dicular  distance  between  the  vertex  and     /'  N\ 

the  base.     In  the  case  of  a  right  cone,     V  J 

the  altitude  is  equal  to  the  length  of  the  ^-^ 

axis;  it  equals  so  in  Figs.  82  and  84.     In  e 

the  case  of  an  oblique  cone,  Fig.  85,  the 

base  is  not  perpendicular  to  the  axis,  and  the  altitude  is  there 

indicated  by  H,  the  perpendicular  distance  between  the  base 

and  the  vertex. 

The  volume  of  any  cane  is  equal  to  one-third  the  product  of 
area  of  the  base  by  the  altitude.  Let  a  =  area  of  base  and  h  =  the 
altitude;  then, 

V  =  \ah  (1) 

If  the  cone  is  a  cone  of  revolution  and  d  =  the  diameter  of 


122        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


the  base,  the  area  of  the  base  is  a  =   .d'z; 
of  a  in  formula  (1), 


substituting  this  value 


V  =  ^0d2h  =  .2618d2A 

1  — 


(2) 


If  the  base  is  an  ellipse,  as  in  Fig.  85,  let  D  =  be,  the  long  diam- 
eter, and  d  =  gf,  the  short  diameter;  then,  the  area  of  the  base  is 

IT 

.Dd  =  a.     Substituting  this  value  of  a  in  formula  (1), 


V  =  ~Ddh  =  .2Gl$Ddh 


(3) 


Example. — Making  no  allowance  for  thickness,  what  is  the  area  of  a 
piece  of  thin  sheet  metal  that  will  just  cover  the  convex  surface  of  a  cone 
with  a  circular  base  having  a  diameter  of  33  in.  and  an  altitude  of  45  in.? 
If  made  of  wood  weighing  48  lb.  per  cubic  foot,  what  is  the  weight  of  the  cone? 

Solution. — The  area  of  the  sheet  metal  will  be  the  same  as  the  area  of 


33 


the  cone;  hence,  using  the  formula  of  Art.  130,  A  =  ir  X  —  X 


>M?) 


=  2484.5  sq.  in.  Ans.  Here  the  radius  is  33  -f-  2,  and  the  slant  height  is 
the  hypotenuse  of  a  right  triangle,  one  leg  of  which  is  the  radius  and  the 
other  leg  is  the  altitude. 

To  find  the  weight,  it  is  first  necessary  to  calculate  the  volume.     Apply- 
ing formula  (2),  above,   V  =  .2618  X  332  X  45  =  12829.5  cu.  in.     Since 

48 
a  cubic  foot  weighs  48  lb.,  the  weight  is  evidently  12829.5  X  r^-fi  =  356.4, 

say  356  lb.     Ans.  1728 

132.  A  frustum  of  a  cone  is  that  portion  of  the  cone  included 

between  the  base  and  a  plane  parallel  to  the  base  and  intersecting 
8  the  cone.      In  Fig.  84,  a'abb'  is 

a  frustum  of  the  cone  s  —ab;  in 
Fig.  85,  abed  is  a  frustum  of 
the  cone  s  —  bgcf.  In  both 
cases,  the  two  bases  are  parallel; 
but  in  Fig.  84,  both  bases  are 
perpendicular  to  the  axis,  while 
in  Fig.  85,  they  are  oblique  to 
the  axis.  It  is  apparent  that  a 
frustum  of  a  right  cone  is  always 
a  right  frustum,  since  if  one 
base  is  perpendicular  to  the  axis 
(as  it  must  be  in  the  case  of  a 

right  cone)  the  other  base  must  also  be  perpendicular  to  the  axis. 
If  a  frustum  of  a  right  cone  be  rolled  on  a  plane  surface,  the 

area  thus  generated  on  the  plane,  called  the  development  of  the 


Fig.  85. 


§2  MENSURATION  OF  SOLIDS  123 

frustum,  will  have  the  outline  caa'c',  Fig.  83,  which  may  be 
considered  to  be  the  development  of  the  frustum  in  Fig.  84. 
Here  sa  corresponds  to  the  element  sa,  Fig.  84,  and  sc  corresponds 
to  sa'.  The  area  caa'c'  is  equal  to  the  sector  saa' —  sector  sec'. 
Letting  R  =  radius  of  lower  base  and  r  =  radius  of  upper  base, 
perimeter  of  lower  base  =  2rR  =  arc  aa';  perimeter  of  upper 
base  =  2irr  =  arc  cc';  then  area  of  frustum  is  equal  to  ane-half  the 
sum  of  the  circumferences  of  its  bases  multiplied  by  the  slant  height 
of  the  frustum.  Let  s  =  the  slant  height  =  a'a,  Fig.  84,  then 
X  =  }<z(2irR  +  2wr)s,  from  which 

A  =  ts(R  +r)  =  ^s(D  +  d) 

When  D  and  d  are  the  diameters  of  the  bases. 

133.  The  volume  of  any  frustum  of  a  cone  is  equal  to  one-third  the 
altitude  multiplied  by  the  sum  of  the  areas  of  the  upper  base,  the 
lower  base,  and  the  square  root  of  the  product  of  the  areas  of  the  bases. 

Thus,  let  a'  =  the  area  of  the  lower  base,  a"  =  the  area  of  the 
upper  base,  and  h  =  the  altitude  =  h  in  Fig.  85;  then, 
V  =  Yzhia!  +  a"  +  V^V7)  (1) 

If  the  bases  are  circles,  let  D  =  the  diameter  of  the  lower 

IT 

base  and  d  =  the  diameter  of  the  upper  base;  then,  a  =  ^  dD2, 

a"  =  -tjs,  and  substituting  in  formula  (1)  and  reducing, 

V  =  .261Sh(D2  +  d2  +  Dd)  (2) 

If  the  bases  are  not  perpendicular  to  the  axis,  they  are  ellipses; 
letting  D'  and  d'  be  the  long  and  short  diameters  of  the  lower  base 
and  D"  and  d"  the  long  and  short  diameters  of  the  upper  base, 

a'  =  \D'd',  a"  =  ^D"d",  and  substituting  in  formula  (1)  and 

reducing,  

V  =  .261Sh(D'd'  +  D"d"  +  VO'  W'il'd")  (3) 

Example. — Find  the  weight  of  the  plater  roll  shown  in  Fig.  86;  all  right 
sections  are  circles;  the  roll  is  solid;  and  it  is  made  of  cast  iron  weighing  450 
pounds  per  cubic  foot. 

Solution. — The  parts  marked  A  and  E  are  cylinders  having  the  same 
diameter,  8  in.,  and  are  therefore  equivalent  to  a  single  cylinder  having  a 
diameter  of  8  in.  and  a  length  of  9  +  14.5  =  23.5  in.  The  part  marked  C 
is  a  cylinder  17  in.  in  diameter  and  36  in.  long;  the  parts  marked  B  and  D 
are  frustums  of  cones  whose  bases  have  diameters  of  12  in.  and  17  in.  and  an 
altitude  of  3  in.  The  volume  of  the  roll  will  be  the  sum  of  the  volumes  of 
the  parts  A,  B,  C,  D,  and  E.     The  volume  of  the  cylinders  may  be  calcu- 


124        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


Iated  by  formula  (2),  Art.  125,  and  the  volume  of  the  frustums  by  formula 
(2),  above 

Volume  of  A  and  E  =  .7854  X8!X  (9  +  14.5)  =  1181  cu.  in. 

Volume  of  C  =  .7854  X  172  X  36  =  8171 

Volume  of  B  and  D  =  .2618  X  3(172  +  12*  +  17  X  12)  X  2  =   1001 

Total  volume  =  10,353  cu.  in. 

10,353         ,     ,  .  ,       ,    .         „  .    10,353 

The  number  of  cubic  feet  is      '    §  ,  and  the  weight  of  the  roll  is    172g 

X  450  =  2696  lb.      Ans.     Since  the  constant  450  is  given  to  only  three 
significant  figures,  the  various  results  were  limited  to  four  significant  figures. 


Fig.  86. 


Example  2. — Referring  to  Fig.  85,  suppose  the  long  and  short  diameters 
of  the  lower  base  are  163^  in.  and  11  in.,  and  the  long  and  short  diameters 
of  the  upper  base  are  6.6  in.  and  4.4  in.;  if  the  altitude  is  8  in.,  what  is  the 
volume  of  the  frustum? 

Solution.— Applying  formula  (3),  above,  V  =  .2618  X  8(16.5  X  11  +6.6 
X  4.4  +  Vl6.5  X  11  X  6.6  X  4.4  )   =  5930  cu.  in.     Ans. 

134.  The  Sphere. — If  a  semicircle  be  revolved  about  its 
diameter,  the  surface  thus  generated  is  called  a  spherical  surface; 
the  solid  that  is  bounded  by  a  spherical  surface  is  called  a  sphere. 
In  Fig.  87(a),  suppose  the  semicircle  ras  to  be  revolved  about  its 
diameter  rs;  it  will  then  generate  a  spherical  surface,  and  the  line 
rs  is  called  the  axis  of  the  sphere.  The  middle  point  o  of  the 
axis  is  the  center  of  the  sphere.  Any  right  line  drawn  from  the 
center  o  and  ending  in  the  spherical  surface,  as  the  line  og,  is  the 
radius  of  the  sphere,  and  it  is  evident  that  all  radii  of  a  sphere  are 
equal,  since  they  are  equal  to  the  radius  of  the  semicircle  that 
generates  the  sphere.  Any  line  that  passes  through  the  center 
and  is  terminated  by  the  spherical  surface,  as  the  line  rs,  is  a 
diameter  of  the  sphere,  and  any  diameter  may  be  taken  as  the  axis. 
The  points  where  the  axis  intersects  the  spherical  surface  are 
called  the  poles;  in  Fig.  87(a),  r  and  s  are  the  poles. 


§2 


MENSURATION  OF  SOLIDS 


125 


If  a  sphere  be  intersected 
by  a  plane,  the  figure  thus 
formed  will  be  a  circle;  and 
if  the  plane  passes  through 
the  center  of  the  sphere,  the 
circle  is  called  a  great  circle ; 
otherwise,  it  is  called  a  small 
circle.  In  (a)  Fig.  87,  rasb, 
rcsd,  resf,  etc.  are  great  cir- 
cles; the  circles  represented 
by  the  lines  kj  and  mn  are 
small  circles.  All  great  cir- 
cles are  equal;  but  a  small 
circle  may  have  any  value 
between  that  of  a  great  circle 
and  0.  Assuming  the  earth 
to  be  a  perfect  sphere,  the 
meridians  of  longitude  are 
all  great  circles,  while  all 
the  parallels  of  latitude  (ex- 
cept the  equator)  are  small 
circles. 

135.  The  area  of  a  sphere 
is  equal  to  the  area  of  four 
great  circles.  Let  r  =  the 
radius  of  a  great  circle  = 
radius  of  the  sphere,  and  d 
—  2r  =  the  diameter;  then, 

4=4*r2=Td*=3.1416d8  (1) 

The  volume  of  a  sphere 
is  equal  to  its  area  multi- 
plied by  one-third  the  radius 
=  4wr2  X  %r,  or 

V=  ^rr3  =  i7rd3  =  .5236rf3     (2) 

Ex  imple. — A  certain  manu- 
facturer Btatea  that  a  can  of  his 
paint  will  cover  100  square  feet; 
how  many  cans  will  be  required 
to  give  two  coats  of  paint  to  a 
sphere   100  in.  in  diameter?     If 


126        ELEMENTARY  APPLIED  MATHEMATICS  §2 

the  sphere  were  hollow  and  its  inside  diameter  were  100  in.  how  many- 
gallons  would  it  hold  ? 

Solution. — Applying  formula  (1)  to  find  the  area,  which  multiply  by  2 
since  two  coats  are  to  be  applied, 

A  =  3.1416  X  1002  X  2  =  62,832  sq.  in.  =  62,832  4-  144  =  436+  sq.  ft. 
Therefore,  the  number  of  cans  of  paint  required  is  436  -r  100  =  4.36,  say 
4%  cans.     Ans. 

To  find  the  number  of  gallons  that  the  sphere  will  hold,  apply  formula 
(2)  to  find  the  volume,  which  divide  by  231,  the  number  of  cubic  inches  in  a 
gallon;  thus, 

V  =  .5236  X  1003  =  523,600  cu.  in.,  and  523600  -^-  231  =  2266^  gal.  Ans. 

136.  If  a  sphere  be  intersected  by  two  parallel  planes,  the  part 
included  between  the  planes  is  called  a  spherical  segment. 
Referring  to  (a),  Fig.  87,  kj  and  mn  represent  parallel  planes,  and 
kamnbj  is  a  spherical  segment.  It  will  be  noted  that  a  spherical 
segment  is  somewhat  similar  to  a  frustum  of  a  cone,  and  has  two 
bases,  kj  and  mn.  If,  however,  one  plane  is  tangent  to  the 
sphere,  as  pq,  the  spherical  segment  has  but  one  base.  In  (a), 
Fig.  87,  rkij  and  rkmnj  are  spherical  segments  of  one  base.  The 
perpendicular  distance  between  the  bases  is  the  altitude  of  the 
spherical  segment;  thus,  il  is  the  altitude  of  the  spherical  segment 
kmnj,  ir  is  the  altitude  of  the  spherical  segment  rkij,  etc. 

The  convex  surface  of  a  spherical  segment  is  called  a  zone; 
thus,  the  convex  surface  of  kmnj  is  a  zone  of  two  bases,  and  the 
convex  surface  of  rkij  is  a  zone  of  one  base.  Note  that  a  zone 
means  a  surface — not  a  solid;  there  is  no  such  thing  as  the  volume 
of  a  zone. 

137.  The  area  of  a  zone  is  equal  to  the  circumference  of  a  great 
circle  of  the  sphere  of  which  the  zone  is  a  part  multiplied  by  the  alti- 
tude of  the  zone.  The  altitude  of  a  zone  is  the  same  as  the  altitude 
of  the  spherical  segment.  Let  r  =  the  radius  of  the  sphere,  d 
=  the  diameter  of  the  sphere,  and  h  =  the  altitude  of  the  zone ; 
then, 

A  =  2-wrh  =  Tdh  (1) 

The  volume  of  a  spherical  segme?it  is  equal  to  half  the  sum  of  its 
bases  multiplied  by  its  altitude  plus  the  volume  of  a  sphere  of  which 
that  altitude  is  the  diameter.  Thus,  letting  the  letters  represent 
the  same  quantities  as  before,  and  letting  rx  and  di  be  the  radius 
and  diameter  of  one  base,  and  r2  and  d2  the  radius  and  diameter  of 
the  other  base, 

V  =  yfriri*  +  r22)h  +  %ph*  =  .5236/?[3(r12  +  r22)  +  h*]         (2) 
Also,  V  =  .1309/?  [3  W  +  d22)  +  ih2]  (3) 


§2  MENSURATION  OF  SOLIDS  127 

If  the  spherical  segment  have  but  one  base,  as  rkij,  Fig.  87(a), 
make  r2  =  0  in  formula  (2),  and 

V  =  .5236M3r!2  +  h2)  =  .1309A(3di2  +  4A2)  (4) 

138.  Referring  to  (b)  and  (c),  Fig.  87,  suppose  a  sphere  to  be 
generated  by  revolving  the  semicircle  racs  about  its  diameter  r.s; 
then,  that  part  of  the  sphere  that  is  generated  by  any  sector  of  the 
semicircle  is  called  a  spherical  sector.  In  (6),  Fig.  87,  the  spher- 
ical sector  raob  is  generated  by  the  sector  aor  of  the  semicircle; 
the  remainder  of  the  sphere,  sboa,  is  also  a  spherical  sector.  In 
(c),  Fig.  87,  aobdoc  is  a  spherical  sector  generated  by  the  sector 
aoc.  The  zone  forming  the  convex  surface  of  the  spherical 
sector  is  called  the  base  of  the  sector.  In  (6),  Fig.  87,  the  zone 
abr  is  the  base  of  the  spherical  sector  raob;  and  in  (c),  Fig.  87,  the 
zone  abdc  is  the  base  of  the  spherical  sector  aobdoc. 

The  volume  of  a  spherical  sector  is  equal  to  the  area  of  the  zone 
that  forms  its  base  multiplied  by  one-third  the  radius  of  the  sphere. 
Since  the  area  of  a  zone  is  2-rrrh,  the  area  of  a  spherical  sector  is 
27rr/j  X  y&r,  and 

2ir 

y  =  ^-rVi  =  2.0944r2h 
o 

Example  1. — Fig.  88  shows  a  tank  28  in.  long,  12  in.  wide,  and  filled 
with  water  to  a  depth  of  16  in.  A  ball  SJ-2  in.  hi  diameter  is  partly  sub- 
merged in  the  water,  the  vertical  depth,  measured  on  the  axis  being  6  in. 
To  what  additional  height,  x,  will  the  water  level  be  raised  by  the  ball? 

Solution. — Let  x  =  the  additional  height  of  the  water  due  to  the  ball; 
this  height  will  be  the  same  as  though  an  amount  of  water  equal  in  volume 
to  that  displaced  by  the  ball  had  been  added  to  the  water  in  the  tank,  and 
this  volume  is  equal  to  the  volume  of  a  spherical  segment  of  one  base  having 
an  altitude  of  6  in.  the  diameter  of  the  sphere  being  8.5  in.  To  find  the 
radius  of  the  base,  let  CBDA  be  a  section  of  the  ball.  Fig.  89,  taken  through 
the  center;  then,  CBDA  is  a  great  circle,  and  CD  is  the  axis  of  the  sphere. 
AB  is  the  water  level  and  is  6  in.  from  the  lowest  point  D  of  the  ball.  AB 
is  also  the  diameter  of  the  base  of  the  spherical  segment  ADB,  and  in  Fig. 
89,  is  the  chord  of  the  arc  ACB.  One-half  of  AB  =  AE  is  the  radius  of  the 
base  of  the  spherical  segment.  Now  applying  the  principle  of  Art.  87, 
AE2  =  r,2  =  CE  X  ED  =  (8.5  -  6)  X  6  =  15.  Substituting  this  value 
of  r,2  in  formula  (4),  Art.  137, 

V  =  .5236  X  6(3  X  15  +  62)  =  254.47  cu.  in. 

which  is  the  volume  of  that  part  of  the  ball  submerged  in  the  water.  The 
amount  of  water  in  the  tank  is  28  X  12  X  16  =  537Q  cu.  in.  Adding  to 
this  the  volume  of  the  spherical  segment,  5376  ~r  254.47  =  5630.47  cu.  in. 
Representing  the  depth  of  the  water  after  the  ball  has  been  placed  in  it  by 
d,  the  volume  occupied  by  the  water  and  the  submerged  part  of  the  ball  is 


128        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


28  X  12  X  d  =  5630.47;  from  which,  d  =  5630.47  -=-  (28  X  12)  =  16.7574- 
in.     Hence,  the  ball  raised  the  water  level  16.7574  -  16  =  .7574  in.  =  x. 

Arts. 

If  only  the  value  of  x  had  been  desired,  all  that  would  be  necessary  to 

ascertain  it,  is  to  divide  the  volume  of  the  spherical  segment  by  the  area  of 


Fig.  88. 

the  bottom  of  the  tank,  since  the  volume  of  the  prism  abcdefgh,  Fig.  88,  must 

be  equal  to  the  volume  of  the  submerged  part  of  the  ball,  and  the  area 

of  the  base  of  the  prism  is  the  same  as 
the  area  of  the  bottom  of  the  tank. 
Hence,  254.47  -r  (28  X  12)  =  .7574  - 
in.,  as  before. 

Example  2. — A  ball  15  ft.  in  diameter 
is  to  be  painted  in  two  colors  in  such  a 
manner  that  the  top  and  bottom  zones 
will  be  of  one  color  and  the  middle  zone 
of  another  color,  all  three  zones  having 
the  same  altitude.  If  the  middle  zone 
is  white  and  the  other  two  black,  which 
color  will  require  the  most  paint,  and  w  hat 
will  be  the  area  of  the  surfaces  covered? 
Solution. — The  altitude  of  each  zone  is  15  ■*■  3  =  5  ft.     By  formula  (1), 

Art.  137,  the  area  of  a  zone  is  irdh;  since  d  and  h  are  the  same  for  all  three 

zones,  their  areas  are  equal,  and 

A  =  3.1416  X  15  X  5  =  235.62  sq.  ft. 


§2 


MENSURATION  OF  SOLIDS 


129 


Therefore,  the  area  to  be  covered  with  the  white  painl  is  235.G2  sq.  ft., 
and  the  area  to  be  covered  with  the  black  paint  ifl  235.62  X  2  =  471.24 

sq.  ft.     A7is. 


EXAMPLES 

(1)  Fig.  90  shows  an  iron  casting.  The  part  marked  C  is  8  in.  in  diam- 
eter and  1  in.  thick;  part  D  is  5  in.  in  diameter;  part  H  is  10  in.  square  and 
1%  in.  thick;  there  are  four  strengthening  webs  marked  A,  which  may  l>e 
considered  as  triangular  prisms,  neglecting  the  small  curve  formed  where 
they  join  the  middle  cylinder. 


Fio.  90. 


Taking  the  weight  of  a  cubic  inch  of  casl  iron  as  .2604  11>.,  what  is  the 
weight  of  the  casting?  The  height  over  all  is  10'  _,  in.,  and  a  hole  1  I •_•  in. 
in  diameter  passes  through  the  center  of  the  casting.  Ana.  95.8  lb. 

(2)  A  wooden  ball  18  in.  in  diameter  has  a  9-inch  hole  through  it,  the  axis 
of  the  hole  coinciding  with  the  axis  of  the  ball;  what  was  the  original  volume 
of  the  ball,  and  what  is  the  volume  after  the  hole  has  been  placed  in  the  ball? 
Note  that  the  part  removed  when  the  hole  was  bored  is  a  cylinder  and  two 
spherical  segments  of  one  base,  both  segments  being  equal 

Ana.  3054—  cu.  in.;  1083+  cu.  in. 

(3)  One  side  of  a  stock  chest  12  ft.  high  is  shaped  like  a  semicircle  with  a 
radius  of  5  ft.  2  in.;  the  side  opposite  is  a  right  line  10  ft.  4  in.  long  and  is 


130 


ELEMENTARY  APPLIED  MATHEMATICS 


§2 


joined  to  the  semicircular  part  by  right  lines  7  ft.  long,  the  angle  between 
the  long  side  and  the  two  short  sides  being  a  right  angle.  If  the  chest  stands 
with  its  axis  vertical,  what  is  its  capacity  in  cubic  feet  when  filled  to  within 
8  in.  from  the  top?  Ans.  1295  cu.  ft. 

(4)  What  is  the  capacity  in  United  States  gallons  of  a  vertical  cylindrical 
stock  chest  11  ft.  10  in.  high  and  14  ft.  inside  diameter,  if  it  is  filled  to 
within  one  foot  of  the  top,  no  allowance  being  made  for  the  displacement  of 
the  agitator?  Ans.  12.475  gal. 

(5)  If  the  consistency  of  the  stock  is  275  pounds  per  1000  U.  S.  gallons  in 
the  last  example,  what  is  the  dry  weight  of  the  contents  of  the  stock  chest? 

Ans.  3431  lb. 

(6)  Find  the  solid  contents  in  cubic  feet  of  the  following  logs: 
24  logs  16  ft.  long,  18  in.  at  the  large  end,  15  in.  at  small  end. 
36  logs  15  ft.  long,  17  in.  at  the  large  end,  13  in.  at  small  end. 
38  logs  12  ft.  long,  15  in.  at  the  large  end,  12  in.  at  small  end. 
28  logs  20  ft.  long,  18  in.  at  the  large  end,  14  in.  at  small  end. 

When  calculating  the  cubical  contents  of  logs,  it  is  customary  to  take  *  =  3, 
because  the  logs  are  seldom,  if  ever,  exactly  round,  and  it  is  not  only  useless 
to  take  x  =  3.1416,  but  the  results  will  be  inaccurate. 

Ans.  339,605  cu.  ft. 

(7)  A  cylindrical  tank  28  ft.  long  and  68  in.  in  diameter  lies  with  its  axis 
horizontal;  how  many  gallons  will  it  hold  when  filled  to  within  7.5  in.  of  the 
top?  Ans.  4969  gal. 


Fig.  91. 


(8)  A  tank  reservoir  that  has  the  general  shape  of  a  frustum  of  a  right 
cone  is  16  ft.  in  diameter  at  base,  14  ft.  9  in.  in  diameter  at  top,  and  is  18  ft. 
high;  how  many  gallons  will  it  hold  when  filled  to  within  14  in.  of  the  top? 

Ans.  23,392  gal. 

(9)  A  pipe  Z%  in.  inside  diameter  discharges  water  at  the  rate  of  5.62 
ft.  per  sec;  how  many  gallons  will  it  discharge  in  one  hour?  The  amount 
discharged  is  evidently  equal  to  the  amount  required  to  fill  a  pipe  of  the 
same  cross-sectional  area  and  having  a  length  equal  to  5.62  X  60  X  60  ft. 

Ans.  967.3  gal. 

(10)  Find  the  volume  of  a  frustum  of  a  cone  having  the  dimensions  indi- 
cated in  Fig.  91.  .   130.72  cu.  ft. 


§2  MENSURATION  OF  SOLIDS  131 

SIMILAR  FIGURES 

139.  Let  ABODE,  Fig.  92,  be  any  polygon.     From  any  point  0 

within  the  polygon,  draw  lines,  called  rays,  to  the  vertexes  of  the 

polygon;  and  on  these  rays,  lay  off  distances  OA',  OB',  OC,  etc. 

*u      *v        *•     OA'       OB'      OC  _, 

in  such  manner  that  the  ratios  ~  .    =  77^-  =  ^^  =  etc.      .tor 

OA        OB       OC 

OA' 
instance,  suppose  ~-j-  =  I;  then,  OA'  =  §  OA,  OB'  =  §  OB,  OC 

=  §  OC,  etc.  Joining  the  points  A',  B',  C,  etc.  by  right  lines, 
another  polygon  A'B'CD'E'  is  obtained  that  is  said  to  be 
similar  to  the  polygon  ABODE.     The  angles  at  the  vertexes  of 


Fig.  92. 


the  two  polygons  that  are  similarly  placed  are  equal  and  the 
corresponding  sides  are  proportional;  that  is,  angle  A  =  angle 

,     M  ,  A'B'      B'C      CD' 

A',  angle  B  =  angle  B,  etc.,  and  -^  =  -j^r  =  -gjy  =  etc. 

Hence,  when  two  polygons  are  so  related  that  their  corresponding 
a?igles  are  equal  and  their  corresponding  sides  are  proportional, 
they  are  similar. 

The  point  0  from  which  the  rays  are  drawn  may  be  taken  any- 
where in  plane  of  the  figure — either  within  or  without  the  polygon ; 
in  fact,  it  may  be  taken  outside  the  plane  of  the  figure,  provided 
the  planes  of  the  two  figures  are  parallel,  as  in  Fig.  93.  Here  the 
figure  ABCDEF  is  similar  to  the  figure  A'B'CD'E'F',  and  the 
planes  of  these  figures  are  parallel.  If  the  vertexes  of  the  polygons 
are  connected  by  lines  A' A,  B'B,  etc.,  they  form  the  edges  of  a 
prismoid.  If  the  rays  OA ,  OB,  etc.  be  extended  sufficiently  fario 
enable  another  prismoid  to  be  formed  and  the  edges  of  the  second 
prismoid  are  proportional  to  the  first,  then  the  two  solids  are^ 
similar. 


132        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


In  the  caso  of  similar  plane  figures  or  similar  solids,  lines  that 
arc  similarly  placed  arc  called  homologous;  thus,  in  similar  poly- 
gons, the  corresponding  sides  are  homologous.     For  instance,  in 


Fig.  92,  .45  and  A'B',  BC  and  B'C",  etc.  are  homologous  sides; 
AC  and  A'C",  EB  and  E'B',  etc.  are  homologous  lines.  In  Fig. 
93  if  there  is  another  solid  similar  to  the  one  there  shown,  as 
the  solid  a'b'c'd'e'f'abcdef,  then  the  edges  A'A  and  a'a,  B'B 
and  b'b,  etc.  are  homologous;  and  lines  similarly  placed,  as  B'A 
and  b'a,  A'D  and  a'd,  etc.  are  homologous  lines. 

140.  Suppose  a  plane  figure  to 
be  irregular  in  shape,  as  A  BCD, 
Fig.  94;  if  from  a  point  within  it.  a 
sufficient  number  of  rays  be  drawn, 
and  points  on  these  rays  be  taken 

.  ..  ,  OA'  OB'  OC 
Buchthat^T  =  OB=OC=etc-> 
then  the  figure  A'B' CD'  will  be 
similar  to  the  figure  ABCD.  Any 
lines  similarly  placed  in  the  two 
figures  will  be  homologous  lines;  thus,  the  lines  DA  and  D'A', 
AC  and  A'C,  etc.  are  homologous  lines. 

141.  Given  any  two  similar  plane  figures,  their  perimeters  are 
proportional  to  the  lengths  of  any  two  homologous  lines;  hence,  if 


§2  MENSURATION  OF  SOLIDS  133 

the  perimeter  of  one  of  two  similar  figures  is  known  and  the  lengths 
of  any  two  homologous  linos  arc  also  known  or  can  be  measured, 
the  perimenter  of  the  other  figure  can  be  found  by  proportion. 
Let  p  and  P  be  the  perimeters  and  I  and  L  be  the  lengths  of  two 
homologous    lines;    then, 

P  :p  =  L  :l 

From  which,  P  =  - -  -  and  p  =  -y— 

i  1j 

All  regular  polygons  and  circles  are  similar;  two  ellipses  are 
similar  when  their  long  and  short  diameters  are  proportional; 
two  circular  segments  are  similar  when  the  central  angles  sub- 
tended by  their  arcs  arc  equal;  etc. 

Example. — A  cylindrical  tank  lies  in  a  horizontal  position  and  is  filled 
with  water  to  a  depth  of  31  in.  If  the  inside  diameter  of  the  tank  is  40  in. 
and  it  is  known  that  the  wetted  perimeter  of  a  tank  G  in.  in  diameter  and 
similarly  filled  is  12.919  in.,  what  is  the  wetted  perimeter  of  the  given  tank? 

Solution. — The  wetted  perimeter  is  the  circular  arc  touched  by  the  water. 

Since  the  two  tanks  are  similarly  filled,  the  depths  are  homologous  lines,  the 

diameters  are  also  homologous  lines,  and  the  wetted  perimeters  are  similar. 

Letting  x  =  the  wetted  perimeter  sought,  x  :  12.919=  40  :  6;  whence,  x  = 

12  919  X  40 

g  a  *u  =  g6  128)  gay  g6^  in      Ans 

142.  The  areas  of  any  two  similar  plane  figures  or  solids  are 
proportional  to  the  squares  of  any  two  homologous  lines.  Since 
all  circles  are  similar,  the  areas  of  any  two  circles  are  to  each  other 
as  the  squares  of  their  radii,  the  squares  of  their  diameters,  the 
squares  of  the  chords  of  equal  arcs,  etc.  In  Fig.  94,  if  the  area  of 
ABCD  be  represented  by  A  and  the  area  of  A'B'C'D'  by  o,  and 
the  lengths  of  the  homologous  lines  AC  and  A'C  are  known  and 
are    represented    by   L    and    I, 

A  :a  =  L2  :  I2 

L2  l* 

from  which,  A  =  a  X  -%,  and  a  =  A  X  jy 

Two  cylinders  are  similar  when  their  altitudes  are  proportional 
to  any  two  homologous  lines  and  the  bases  are  similar;  two  cones 
are  similar  when  the  base  of  one  is  similar  to  the  base  of  the 
other  and  the  altitudes  are  proportional  to  any  two  homologous 
lines;  two  prisms  are  similar  when  the  bases  are  similar  and  the 
altitudes  are  proportional  to  any  two  homologous  edges  or  lines 
of  the  prisms;  etc.  Therefore,  if  A  is  the  area  of  the  surface 
of  a  solid  and  a  is  the  area  of  a  similar  solid  (either  the  convex 
area  or  the  entire  area  of  the  solids)  and  L  and  I  are  the  lengths 


134        ELEMENTARY  APPLIED  MATHEMATICS  §2 

of  any  two  homologous  lines  (the  diameter  or  altitude  of  a 
cylinder,  prism,  or  cone,  or  the  diameter  of  either  base  of  a  fru- 
stum of  a  cone,  or  the  length  of  any  edge  of  a  prism  is  a  homo- 
logous line  when  compared  with  a  similar  solid),  then 

A  :a  =  I2  :l2 

Example. — Referring  to  the  example  of  Art.  131,  what  is  the  area  of  a 
piece  of  sheet  metal  that  will  exactly  cover  a  cone  having  a  circular  base 
that  is  26.4  in.  in  diameter  and  whose  altitude  is  36  in.? 

Solution. — Since  the  area  of  a  cone  having  a  diameter  of  base  of  33  in. 

and  an  altitude  of  45  in.  is  2484.5  sq.  in.,  and  since  -=5-  =  j^  =  .8,  the  cones 
are  similar,  and  letting  A  =  the  area  of  the  cone  here  being  con- 
sidered,  A :  2484.5  =  362:  452,  from  which  A  =  2484.5  X^  =  2484.5  i^J 

=  2484.5  (.)  2=  2484.5  X  .S2  =  1590.1  -  sq.  in.     Ans. 
If  the  area  had  been  calculated  by  the  formula  of  Art.  130, 

x  ^^  x  \j'SQ2  +  (— )  '  =  1590.1  -,  the  same  result  as  before. 

143.  The  volumes  of  any  two  similar  solids  are  to  each  other  as  the 
cubes  of  their  homologous  lines.  Let  V  =  the  volume  of  one  solid 
and  v  =  the  volume  of  a  similar  solid;  then,  if  L  and  I  are 
homologous  lines, 

V  :  v  =  L3  :  I3 

L3  iJ  \ 

from  which,  V  =  v  X  -y  =  v  \~n) ' 

144.  Any  cylinder  or  prism,  any  pyramid  or  cone,  any  frustum 
of  a  pyramid  or  cone,  and  any  sphere  or  spherical  segment  may 
be  regarded  as  a  prismoid,  and  its  volume  may  be  calculated  by 
the  prismoidal  formula.  For  instance,  in  the  case  of  a  cone,  one 
base  of  the  prismoid  is  0;  hence,  comparing  the  middle  section 
with  the  base,  a  homologous  line  of  the  base  will  be  twice  as 
long  as  the  corresponding  homologous  line  of  the  middle  section, 
and  the  area  of  the  base  will  be  22  =  4  times  the  area  of  the 
middle    section.    Letting   a  =  area    of   the    base,  the  area  of 


A  = 


a 


the  middle  section  is  j;  and    by    the    prismoidal    formula,     Y 

=  lh  (a  +  4  X  4  +  O)  =  lah,  which  is  the  same  as  formula  (1), 

Art.  131.  In  the  case  of  a  sphere,  both  ends  of  the  prismoid  are 
points  and  their  areas  are  0;  the  area  of  the  middle  section  is 
7rr2  and  the  altitude  is  2r.  Therefore,  by  the  prismoidal  formula, 
V  =  I  X  2r(0  +  4  X  rrr2  -f  0)  =  frrr3,  which  is  the  same 
as  formula  (2),  Art.  135. 


§2  MENSURATION  OF  SOLIDS  135 

If  the  cross  sections  of  one  prismoid  are  proportional  to  similar 
cross  sections  of  another  prismoid,  but  the  altitudes  of  the  prismoids 
are  not  in  the  same  proportion,  the  volumes  of  the  prismoids  are 
proportional  to  the  products  of  the  areas  of  the  cross  sections  by 
the  altitudes.  For  instance,  suppose  that  the  dimensions  of  the 
frustum  of  two  cones  are  known,  that  the  cross  sections  of  one 
frustum  are  proportional  to  the  cross  sections  similarly  placed 
in  the  other  frustum,  but  that  the  ratio  of  the  altitudes  (or  of 
any  homologous  lines  not  in  the  section  or  in  planes  parallel  to 
the  sections)  is  not  the  same  as  the  ratio  of  the  cross  sections; 
then  letting  A  and  a  be  the  areas  of  any  two  homologous  sections 
and  H  and  h  the  altitudes,  V  :  v  =  A  X  H  :  a  X  h.  Further, 
letting  L  and  I  be  any  two  homologous  lines  in  the  sections  whose 

areas  are  A  and  h,  —  =   l-y)  ■     From  the  foregoing  proportion, 
V  =  v  ( —  j  -,-  •     Substituting  the  value  of  — 


and 


r->m 


-*m 


145.  Suppose  that  through  any  point  0  in  any  solid,  three 
planes  are  passed  perpendicular  to  one  another,  as  the  planes 
AB,  CD,  and  EF,  Fig.  95;  these  planes  intersect  in  the  right 
lines  mn,  pq,  and  rs.  If  now  measurements  are  taken  in  these 
planes  parallel  to  the  intersecting  lines  and  these  measurements 
are  proportional  to  similar  measurements  taken  in  another  body, 
and  this  proportion  is  the  same,  no  matter  where  the  point  0  may 
be  situated,  the  two  solids  are  similar,  and  their  volumes  are  pro- 
portional to  the  products  of  any  set  of  three  homologous  lines. 
For  instance,  suppose  the  point  0  be  located  at  a  corner  of  the  base 
of  a  rectangular  parallelopiped;  then,  a  measurement  parallel  to 
rs,  mn,  and  pq,  will  correspond  to  the  length,  breadth,  and  thick- 
ness of  the  parallelopiped,  which  may  be  represented  byL,  B,  and  T, 
respectively.  Homologous  lines  of  a  similar  parallelopiped  may 
be  designated  by  I,  b,  and  t,  and 

V   :  v  =  L  X  B  X  T  :  iXb  Xt 

Example. — Referring  to  example  2,  Art.  133,  what  is  the  volume  of  a 
similar  frustum  of  a  cone  having  an  altitude  of  13 y>  in.? 

Solution. — Since  the  frustums  are  similar,  the  volumes  of  the  two  frus- 


136        ELEMENTARY  APPLIED  MATHEMATICS 


|2 


turns  are  proportional  to  the  cubes  of  any  two  homologous  lines,  Bay  their 

/13.5\  s 
altitudes;  then.  V:  v  =  H3:h3,or  P:5930  =  13.53:83,  and  V  =  5930 (-g    1 

=  28497  cu.  in.     Ans. 


Fig.  95. 

Example  2. — Suppose  that  the  bases  of  the  frustums  in  the  preceding 
example  had  been  similar,  the  diameters  of  the  upper  bases  being  11  in. 
and  14>£  in.,  while  the  altitudes  are  8  in.  and  12^  in.  respectively;  what  is 

the  volume  of  the  latter  frustum? 

Solution. — The  ratio  of  the  bases 
(and  of  any  section  parallel  to  the 
bases  is  14.25  -5-  11  =  1.3  — ;  the  ratio 
of  the  altitudes  is  12.375  -h  8  =  1.5  4- ; 
hence,  the  volumes  are  proportional 
to  the  bases  multiplied  by  the  alti- 
tudes, and  V  =  v  (-A  r-  =  5930 
x  (^)  ■  x    12375 

Ans. 

Fig.  96.  Example  3. — Referring  to  the  ex- 

ample of  Art.  141,  suppose  the  40-inch 
tank  is  14  ft.  5  in.  long  and  the  6-inch  tank  is  3  ft.  8  in.  long;  how  many 
gallons  does  each  hold  when  filled  to  the  depth  specified? 

Solution.— A  cross  section  of  the  40-inch  tank  is  shown  in  Fig.  96. 
The  wetted  perimeter  is  the  length  of  the  arc  ACB,  which  was  found  to  be 


=  15394  cu.  in. 


§2 


MENSURATION  OF  SOLIDS 


137 


8C.127  in.  long.    The  area  of  the  segment  ACB  is  equal  to  the  area  of  the 

sector  AC  BO  +  area  of  triangle  AOB.  Using  the  principle  of  Art.  87, 
AD  =  y/9  X  31  =  16.703  in.  Area  of  sector  =  y2  x  861.27  X  20  =  86.127 
sq.  in.  Area  of  triangle  =  16.703  X  (20  -  9)  =  183.73  sq.  in.  Ana  of 
segment  =  861.27  +  183.73  =  1045  sq.  in.  Hence,  volume  of  water 
in  40-inch  tank  is,  since  14  ft.  5  in.  =  173  in.,  1045  X  173  =  180785  cu.  in. 
-  180785  -r  231  =  782.6+ gal.  Ans. 
To  find  the  volume  of  the  water  in  the  6-inch  tank,  use  the  proportion 

44 


given  above,  and  782.6:  v 
4.4785—  gal.     Ans. 


402  X  173  :62  X  44,  or  v  =  782.6 


o 


/. 


173 


SYMMETRICAL  FIGURES 

146.  Two  figures  are  said  to  be  symmetrical  with  respect  to 
a  line,  called  the  axis  of  symmetry,  when  any  perpendicular 
to  the  axis  that  is  limited  by  the  outline  of  the  figure  is  bisected 
by  the  axis.  Thus,  referring  to  Fig.  97,  if  the  perpendiculars 
A' A,  B'B,  CD,  etc.  are  bisected  by  the  line  mn,  mn  is  an  axis  of 
symmetry,  and  the  two  figures 
ABODE  A  and  A'B'C'D'E'A'  are 
symmetrical.  Now  suppose  that 
FBCDG  and  FB'C'D'G  are  sym- 
metrical with  respect  to  the  axis  mn, 
the  sides  FG  coinciding;  then  the 
figure  FBCDGD'C'B'F  is  said  to  have 
an  axis  of  symmetry. 

If  two  figures  are  symmetrical,  they 
are  evidently  equal;  thus,  in  Fig.  97, 
ABODE  =  A'B'C'D'E'.  Conse- 
quently, if  any  figure  have  an  axis  of 
symmetry,  this  axis  bisects  the  figure; 
hence,  FBCDG  =  FB'C'D'G'  =  one- 
half  FBCDGD'C'B'F.  It  is  also 
evident  that  if  the  plane  of  the  paper  be  folded  on  the  line  mn, 
A  will  fall  on  A',  B  will  fall  on  B',  0  will  fall  on  (',  etc,  ami 
ABODE  will  be  superposed  on  and  will  coincide  with  A'B' 
C'D'E'.  Also,  the  left  half  of  FBCDGD'C'B'F  will  be  superposed 
on  and  will  coincide  with  the  right  half;  therefore,  if  a  plan, 
figure  can  be  so  folded  that  every  point  on  one  side  of  the  line  of 
folding  will  coincide  with  a  point  on  tin-  other  side,  the  line  of  folding 
will  be  an  axis  of  symmetry  and  will  divide  the  figure  into  two  equal 
parts. 


i) 

a/ 

F 

\A' 

ff 

c 

c 

\L 

sk/ 

D 

z>' 

Ik..   '.•: 


138 


ELEMENTARY  APPLIED  MATHEMATICS  §2 


147.  If  the  figure  have  two  axes  of  symmetry,  as  mn  and  pq,  Fig. 
98,  their  point  of  intersection  0  is  called  the  center  of  symmetry, 
and  the  center  of  symmetry  is  the  geometrical  center  of  the  figure. 
Every  regular  polygon  has  a  center  of  symmetry,  which  may  be 


Fig.  98. 

found  by  drawing  lines  from  the  vertexes  to  the  opposite  vertexes, 
when  the  polygons  have  an  even  number  of  sides,  as  4,  6,  8,  etc. ; 
but,  if  the  polygons  have  an  odd  number  of  sides,  the  axes  of 
symmetry  are  found  by  drawing  lines  from  the  vertexes  perpen- 
dicular to  the  sides  opposite  them.     If  the  number  of  sides  is  even, 


Fig.  99. 

a  line  drawn  perpendicular  to  any  side  at  its  middle  point  will 
bisect  the  opposite  side  and  be  an  axis  of  symmetry.  Conse- 
quently, a  regular  polygon  has  as  many  axes  of  symmetry  as  it  has 
sides.  See  (A)  and  (B),  Fig.  99.  The  figure  may  be  folded  on  any 
one  of  these  axes  and  one  half  will  coincide  with  the  other  half. 


§2  MENSURATION  OF  SOLIDS  139 

An  isosceles  triangle  and  an  arc,  sector,  or  segment  of  a  circle 
has  but  one  axis  of  symmetry;  see  (D),  Fig.  99.  The  figure 
whose  outline  is  shown  at  (C)  also  has  but  one  axis  of  symmetry. 
An  ellipse  has  two  axes  of  symmetry,  the  long  and  short  diameters. 

148.  A  solid  has  a  plane  of  symmetry  when  sections  equally 
distant  from  the  plane  of  symmetry  and  parallel  to  it  are  equal 
and  every  point  in  one  section  has  its  symmetrical  point  in  the 
other  section.  A  f  rust  rum  of  any  cone  has  at  least  one  plane  of 
symmetry,  which  includes  the  axis  of  the  cone  and  the  long 
diameter  of  either  base.  If  the  frustum  is  that  of  a  right  cone, 
the  bases  are  perpendicular  to  the  axis,  and  there  are  any  num- 
ber of  planes  of  symmetry  perpendicular  to  the  bases. 

If  a  solid  has  two  planes  of  symmetry,  they  intersect  in  a  line  of 
symmetry,  and  if  it  has  three  planes  of  symmetry,  one  of  which  is 
at  right  angles  to  the  other  two,  the  three  planes  intersect  in  a 
point  of  symmetry,  which  is  the  center  of  the  solid.  Thus,  in 
Fig.  95,  if  the  plane  AB  is  a  plane  of  symmetry,  that  part  of  the 
rectangular  parallelopiped  in  front  of  the  plane  is  symmetrical  to 
that  part  behind  it;  if  the  plane  EF  is  also  a  plane  of  symmetry, 
that  part  of  the  solid  to  the  right  of  the  plane  is  symmetrical  to 
that  part  to  the  left,  and  the  two  planes  intersect  in  the  line  of 
symmetry  pq.  If  a  third  plane  AB,  perpendicular  to  the  other 
two  is  also  a  plane  of  symmetry,  that  part  of  the  solid  below  the 
plane  is  symmetrical  to  that  part  above  it,  and  the  three  planes 
intersect  in  the  point  0,  which  is  the  center  of  the  solid. 

149.  Now  observe  that  when  a  body  has  one  plane  of  sym- 
metry, the  plane  divides  the  body  into  two  equal  parts,  and  the 
center  of  the  body  lies  somewhere  in  this  plane.  When  the  body 
has  two  planes  of  symmetry,  the  center  lies  in  both  planes  and, 
hence,  lies  somewhere  in  the  line  of  their  intersection;  the  two 
planes  divide  the  body  into  four  parts,  and  if  they  are  perpen- 
dicular to  each  other,  they  divide  the  body  into  four  equal  parts. 
When  the  body  has  three  planes  of  symmetry,  the  center  lies  in  all 
three  planes,  which  have  only  one  common  point— the  point  of 
intersection;  the  three  planes  divide  the  body  into  eight  equal 
parts,  and  if  the  planes  are  perpendicular  to  one  another,  the 
eight  parts  are  all  equal. 

To  find  the  center  of  a  line,  bisect  the  line  and  draw  a  right  line 
perpendicular  to  the  given  line  at  the  point  of  bisection;  this  line 
will  be  an  axis  of  symmetry,  provided  the  line  is  symmetrical 


140        ELEMENTARY  APPLIED  MATHEMATICS 


§2 


(as  in  the  case  of  a  right  line  or  circular  arc) ;  otherwise,  it  has  no 
center,  unless  it  is  symmetrical  with  respect  to  a  point.  To  be 
symmetrical  with  respect  to  a  point,  every  line  drawn  through  the 
point  and  limited  by  the  given  line  or  by  the  perimeter  of  the 
given  figure  must  be  bisected  by  the  point,  which  is  called  a 
center  of  symmetry.  Thus,  the  line  shown  in  Fig.  100  at  (a)  is 
symmetrical  with  respect  to  the  center  0,  because  OA  =  OA', 


Fig.  100. 

OB  =  OB',  OC  =  OC,  etc.,  and  0  is  the  center  of  the  line.  For 
the  same  reason,  the  parallelogram  at  (b)  is  symmetrical  with 
respect  to  the  center  0.  The  line  shown  at  (c)  is  not  symmetrical 
with  respect  to  a  center  or  to  an  axis,  and  therefore  has  no  center. 
Observe  that  neither  the  line  at  (a)  nor  the  parallelogram  at  (b) 
have  an  axis  of  symmetry — neither  can  be  folded  on  any  line  so 
one-half  can  be  superposed  on  the  other ;  but  both  have  a  center  of 
symmetry,  which  is  the  center  of  the  figure. 


SOLIDS  OF  REVOLUTION 

160.  Center  of  Gravity. — The  center  of  gravity  of  a  plane 
surface  or  section  is  that  point  at  which  the  surface  will  balance. 
If  the  surface  have  an  axis  of  symmetry,  the  center  of  gravity 
(which  may  be  denoted  by  the  abbreviation  c.  g.)  will  lie  in  that 
axis;  and  if  it  have  two  axes  of  symmetry,  the  c.  g.  will  be  their 


§2 


MENSURATION  OF  SOLIDS 


141 


point  of  intersection.  The  practical  way  of  determining  the 
center  of  gravity  of  a  section  is  to  make  a  scale  drawing  of  it  on 
stiff  paper,  say  Bristol  board  or  cardboard,  then  carefully  cut  out 
the  section;  next  suspend  the  model  against  a  vertical  surface; 
from  a  horizontal  pin  or  needle  passed  through  the  model;  suspend 
the  line  of  a  plumb  bob  from  the  same  pin,  and  where  the  line 
crosses  the  model,  draw  a  line.  Now  suspend  the  model  from 
some  other  point  and  draw  a  line  where  the  line  of  the  plumb  bob 
crosses;  the  intersection  of  the  two  lines  will  be  the  center  of 


Fio.  101. 


gravity  of  the  plane  surface  or  section.  This  method  of  deter- 
mining the  c.  g.  is  clearly  shown  in  P'ig.  101.  The  model  is  first 
drawn  and  then  cut  out,  including  the  holes;  a  pin  hole  is  made  at. 
A,  so  that  the  heavy  part  of  the  model  will  hang  lowest;  where 
the  plumb  line  crosses,  the  line  mn  is  drawn.  The  model  is  then 
suspended  from  another  point  B  and  another  plumb  line  drawn, 
which  crosses  mn  at  0,  and  0  is  the  c.  g.  of  the  section. 

If  the  section  have  one  axis  of  symmetry,  it  is  necessary  to 
draw  but  one  plumb  line,  since  where  this  crosses  the  axis  of 
symmetry  will  be  the  c.  g.  It  is  advisable  to  have  the  plumb 
lines  cross  at  as  nearly  a  right  angle  as  possible,  since  the  point  of 
intersection  will  then  be  easier  to  determine. 

If  the  section  (model)  be  placed  on  a  sharp  point  directly  under 


142        ELEMENTARY  APPLIED  MATHEMATICS  §2 

the  point  0,  it  will  balance;  if  laid  on  a  knife  edge  along  either  of 
the  two  plumb  lines,  it  will  balance. 

151.  If  any  plane  section  be  revolved  about  a  line  in  that 
section  as  an  axis,  the  solid  so  generated  is  a  solid  of  revolution. 
Some  examples  of  solids  of  revolution  are  the  right  cylinder,  right 
cone,  and  the  sphere.  The  hollow  cylinder  or  tube  of  Art.  127 
is  also  a  solid  of  revolution,  and  may  be  generated  by  revolving  a 
rectangle  about  an  axis  parallel  to  one  of  the  sides  of  the  rectan- 
gle; see  (a),  Fig.  102.  If  a  circle  be  revolved  about  an  axis,  the 
resulting  solid  will  be  what  is  called  a  cylindrical  ring  or  torus; 
see  (6),  Fig.  102.  If  the  revolving  section  is  a  flat  ring,  as  indi- 
cated by  the  dotted  circle  in  (6),  the  torus  will  be  hollow. 


i_  i 

tr-r 


Fig.    102. 

Whatever  the  shape  of  the  revolving  section,  the  volume  of  the  re- 
sulting solid  will  be  equal  to  the  area  of  the  revolving  {generating) 
section  multiplied  by  the  distance  passed  through  by  the  center  of 
gravity  of  the  section.  If  r  =  the  perpendicular  distance  from  the 
center  of  gravity  to  the  axis,  and  the  section  make  a  complete 
revolution,  the  distance  passed  through  by  the  center  of  gravity 
will  be  the  circumference  of  a  circle  having  a  radius  r;  representing 
this  circumference  by  c,  c  =  2-n-r.  If  a  =  the  area  of  the  gen- 
erating section,  the  volume  of  the  solid  is 


I" 


2rra 


152.  Referring  to  (a),  Fig.  102,  the  center  of  gravity  O  of  the 
section  lies  midway  between  AD  and  BC,  because  the  rectangle 
has  two  axes  of  symmetry,  one  of  which  is  parallel  to  and  half 
way  between  AD  and  BC;  hence,  if  the  diameter  of  the  circle 
described  by  O  be  represented  by  dm,  and  AB  and  AD  be  repre- 
sented by  t  and  I,  respectively,  the  area  of  the  rectangle  is  t  X  I, 
and  V  =  wdmtl,  which  is  the  same  as  formula  (2),  Art.  127. 


§2  MENSURATION  OF  SOLIDS 


143 


Referring  to  {b),  Fig.  102,  the  c.  g.  of  the  section  is  the  center  of 
the  generating  circle.  Let  r,  =  the  radius  of  this  circle;  then  the 
area  of  the  generating  section  is  wrf,  and  the  volume  of  the  torus 
is 

V  =  2w2r^r  =  19.7392r!2r  (1) 

If  the  torus  is  hollow,  let  n  =  radius  of  outer  circle  of  section 
and  r2  =  radius  of  inner  circle  of  section;  the  area  of  the  section 
is  7r(r2!  -  r22)  =  7r(ri  +  r2)  (r,  -  rs),  and  V  =  2rr  X  ir(r,  4-  r2) 
(>'i  -  r%),    or 

V  =  19.7392(r14-r2)(r1  -  r2)r  (2) 

Many,  if  not  the  majority,  of  solids  of  revolution  that  occur  in 
practice  have  generating  sections  that  have  an  axis  of  symmetry 
perpendicular  to  the  axis  of  revolution;  thus,  referring  to  (c), 
Fig.  102,  the  sectionA£CZ>  has  an  axis  of  symmetry  O'B  perpen- 
dicular to  the  axis  mn  about  which  the  section  is  revolved.  The 
e.g.  must  lie  somewhere  on  O'B,  and  may  be  located  by  making  a 
model  section,  on  which  draw  O'B,  and  then  suspending  the 
model  from  a  pin  as  previously  described,  indicate  where  the 
plumb  line  crosses  O'B.  Or,  if  preferred,  the  model  may  be 
balanced  on  a  knife  edge,  and  where  this  edge  crosses  O'B  will 
be  the  point  0,  the  e.g. 

Methods  of  finding  by  calculation  the  center  of  gravity  of 
various  geometrical  figures  will  be  described  in  the  text  treating 
on  mechanics. 

Example. — What  is  the  volume  of  a  torus,  if  the  diameter  over  all  is  13>£ 
in.  and  the  diameter  of  a  radial  section  is  %  in.?  What  is  its  weight  if  made 
of  cast  iron,  a  cubic  inch  of  which  weighs  .2604  lb.? 

Solution.— Referring  to  (6),  Fig.  102,  O'A  =  13.5  ■*■  2  =  6.75  in.; 
AB  =  %  in.;  OA  =  %  4-  2  =  Ke  =  -4375  in.  =  r,  in  formula  (1).  O'O 
=  6.75  -  .4375  =  6.3125  in.  =  r;  then, 

V  =  19.7392  X  .43752  X  6.3125  =  23.85-  cu.  in.     Ans. 

The  weight  =  23.85  X  .2604  =  6.211-  lb.     Ana. 

By  radial  section  is  meant  a  section  made  by  a  plane  passing 
through  the  center  of  the  body;  such  a  section  includes  the  axis 
of  the  body.  All  the  sections  shown  in  Fig.  102  are  radial 
sections. 


ELEMENTARY  APPLIED 
MATHEMATICS 

(PART  3) 


EXAMINATION  QUESTIONS 

(1)  Find  (a)  the  convex  area  and  (&)  the  entire  area  of  a  cone 
of  revolution  whose  base  is  22  in.  in  diameter  and  whose  altitude 
is  38  in.  (  (0)   1367.1  sq.  in. 

I  (6)    1747.2  sq.  in. 

(2)  A  hopper  having  somewhat  the  shape  of  a  frustum  of  a 
pyramid  has  the  following  dimensions:  upper  end,  a  rectangle 
44  in.  by  28  in.;  lower  end,  a  square  8  in.  by  8  in.;  altitude  36  in. 
How  many  gallons  will  it  take  to  fill  the  hopper? 

Ans.  82.286  gal. 

(3)  A  piece  of  cast  iron  has  the  shape  of  a  pentagonal  prism; 
the  ends  are  regular  pentagons,  each  edge  measuring  2\  in.,  and 
the  altitude  is  A.\  in.  What  is  its  weight,  a  cubic  inch  of  cast 
iron  weighing  .2604  lb.?  Ans.  10.206  lb. 

(4)  The  base  of  a  triangular  pyramid  is  an  equilateral  triangle, 
one  edge  which  measures  h\  in.;  the  altitude  is  9|  in.;  what  is  its 
volume?  Ans.  43.116  cu.  in. 

(5)  A  shallow  pan  has  the  shape  of  a  frustum  of  a  cone,  and 
its  inside  dimensions  are:  upper  end  14  in.  diameter,  lower  end 
13  in.  diameter,  and  distance  between  ends  2|  in.;  what  are  the 
cubical  contents  of  the  pan?  Ans.  304.31  cu.  in. 

(6)  The  base  of  a  wrought-iron  wedge  is  10|  in.  by  3|  in.;  the 
upper  edge  is  1\  in.  and  the  altitude  is  15  in.  Taking  the  weight 
of  a  cubic  inch  of  wrought  iron  as  .2778  lb.,  what  is  the  weight 
of  the  wedge?  Ana.  64.328  lb. 

(7)  How  many  cubic  yards  are  contained  in  a  foundation 
wall  whose  length  is  24  ft.  3  in.,  breadth  is  15  ft,  8  in.,  inside 
measurements,  and  thickness  is  16  in.?  The  height  of  the  wall  is 
8  ft.  6  in.  A  ns.  365.88  cu.  yd. 

io  145 


146        ELEMENTARY  APPLIED  MATHEMATICS  §2 

(8)  A  wooden  ball  has  a  cylindrical  hole  extending  through  it, 
the  axis  of  the  hole  coinciding  with  the  axis  of  the  ball.  If  the 
diameter  of  the  ball  is  11  in.  and  of  the  hole  is  A\  in.,  what  is 
the  weight  of  the  ball,  if  a  cubic  inch  of  the  wood  weighs  .0231b.? 

Ans.  12.268  1b. 

(9)  A  cylindrical  tank  lies  with  its  axis  in  a  horizontal  position; 
it  is  filled  with  stock  to  within  8  in.  of  the  top;  its  diameter  is 
60  in.,  and  its  length  is  21  ft.  6  in.,  inside  measurements.  How 
many  gallons  of  stock  are  in  the  tank?  Ans.  2907.6  gal. 

(10)  A  cylindrical  tank  48  in.  in  diameter  and  16  ft.  long  is 
partly  filled  with  stock.  The  tank  lies  in  such  position  that 
the  level  of  the  liquid  just  touches  the  upper  end  of  the  dia- 
meter at  one  end,  and  cuts  the  diameter  at  the  other  end  in  its 
middle  point.     How  many  gallons  of  stock  are  in  the  tank? 

Ans.  319.17  gal. 

(11)  It  is  desired  to  make  a  cylindrical  tank  to  hold  800  gal., 
the  height  to  be  1|  times  the  diameter;  what  should  be  the  in- 
side diameter  and  height  to  the  nearest  16th  of  an  inch? 

.  /  Diameter,  53  ic  in. 

'    I  Height,  80|  in. 

(12)  A  wrought-iron  shell  9  ft.  1\  in.  long  is  open  at  both  ends; 
it  is  54  in.  outside  diameter  and  f  in.  thick.  Taking  the  weight 
of  a  cubic  inch  of  wrought  iron  as  .2778  lb.,  what  is  the  weight 
of  the  shell?  Ans.  3362.7  lb. 

(13)  If  the  periphery  of  an  ellipse  whose  diameters  are  37  in. 
and  12^  in.  is  73.22  in.,  what  is  the  periphery  of  an  ellipse  whose 
diameters  are  9j  in.  and  3|  in.?  Ans.  18.305  in. 

(14)  The  area  of  a  certain  figure  is  430.6  sq.  in.,  and  the  length 
of  a  line  drawn  through  the  figure  is  21  j  in.;  what  is  the  area  of 
a  similar  figure,  if  the  length  of  a  line  similarly  placed  is  19f  in.? 

Ans.  357.96  sq.  in. 

(15)  The  volume  of  a  frustum  of  a  cone  having  an  altitude  of 
8y  in.  is  6473  cu.  in.;  what  is  the  volume  of  a  similar  cone  having 
an  altitude  of  6f  in.?  Ans.  3352  cu.  in. 

(16)  What  is  the  volume  generated  by  revolving  an  isosceles 
triangle  about  an  axis  parallel  to  the  base,  under  the  following 
conditions?  base  is  6£  in.,  the  other  two  sides  are  each  4£  in., 
distance  from  axis  to  base  is  7  in.  The  distance  from  the  center 
of  gravity  of  any  triangle  to  the  base  is  one-third  the  altitude. 

Ans.  513.84  cu.  in. 


SECTION  3 

HOW  TO  READ  DRAWINGS 


REPRESENTING    SOLIDS   ON   PLANES 


PRELIMINARY  EXPLANATIONS 

1.  The  Picture  Plane. — When  viewing  an  object,  it  is  essential 
that  the  object  be  illuminated  in  some  manner,  either  by  light 
originating  in  or  on  the  object  itself  or  by  light  coming  from  an- 
other source  and  being  reflected  from  the  object.  Light  travels  in 
right  (straight)  lines,  called  rays,  and  no  matter  what  their  length, 
whether  a  fraction  of  an  inch  or  millions  of  miles,  every  ray  of 
light  is  absolutely  straight.  The  number  of  rays  of  light  from 
any  particular  object  is  infinite,  and  they  extend  in  every  direc- 
tion (in  right  lines)  from  the  object  unless  they  are  stopped 
by  some  opaque  substance  that  light  cannot  penetrate.  A 
certain  number  of  these  rays  enter  the  eye,  wherein  an  image  or 
picture  of  the  object  is  formed  and  the  object  is  then  said  to 
be  seen. 

If,  when  viewing  an  object,  a  sheet  of  paper  (transparent  so 
light  can  pass  through  it)  be  held  between  the  eye  and  the  object 
and  the  various  lines  seen  on  the  object  are  traced  on  the  paper 
with  a  pen  or  pencil,  the  result  will  be  a  drawing  or  picture  of  the 
object  viewed  from  the  position  occupied  by  the  eye  relative  to 
the  object.  For  every  change  in  the  position  of  the  eye,  there  will 
be  a  change  in  the  shape  of  the  drawing,  and  for  every  change 
in  distance  between  the  paper  and  the  eye,  there  will  be  a 
change  in  the  size  of  the  drawing,  these  two  facts  are  clearly 
shown  in  Fig.  1.  Here  ABCDEFG  is  a  rectangular  prismoid — in 
this  case,  a  frustum  of  a  rectangular  pyramid — which  contains  a 
hole  pasing  part  way  through  it;  S  is  the  eye;  and  P  is  the 
plane  of  the  paper,  called  the  picture  plane.  Now  imagine 
lines  drawn  from  the  points  A,  B,  C,  etc.  of  the  object  to  the 
§3  X 


HOW  TO  READ  DRAWINGS 


§3 


eye;  these  lines  correspond  to  the  rays  of  light  from  the  object 
to  the  eye,  and  they  pierce  the  picture  plane  P  in  the  points 
a,  b,  c,  etc.  Joining  these  points  by  lines  as  shown,  the  result  is 
the  outline  abcdefg,  which  is  a  drawing  or  picture  of  the  object 
ABCDEFG.  It  is  evident  that  if  the  point  S  be  moved  up  or 
down  the  line  S'S"  or  be  moved  perpendicular  to  this  line  in  a 
plane  perpendicular  to  the  plane  of  the  paper,  say  in  a  plane 
parallel  to  the  picture  plane,  the  rays  will  make  a  different 
angle  with  the  picture  plane  and  the  shape  of  the  drawing  will 


Fiu.   1. 


be  different.  It  is  further  evident  that  as  the  picture  plane 
is  brought  nearer  to  the  eye,  the  drawing  will  be  smaller,  and  if 
moved  farther  fiom  the  eye,  the  drawing  will  be  larger,  owing 
to  the  convergence  of  the  rays. 

2.  Note  that  every  ray  makes  an  angle  with  the  picture  plane 
that  is  different  from  that  made  by  any  other  ray;  thus,  the  angle 
made  by  the  ray  AS  is  different  from  that  made  by  BS,  FS,  etc. 
Now  assume  the  picture  plane  to  be  so  located  that  it  is  perpen- 
dicular to  the  parallel  planes  of  the  bases  of  the  frustum  and  is 
parallel  to  the  parallel  edges  CB  and  FG;  assume  further  that  the 
point  S,  called  the  point  of  sight,  lies  on  a  ray  passing  through  the 
center  of  the  face  CBGF  and  perpendicular  to  the  picture  plane. 
Then,  if  the  picture  plane  be  very  near  the  object  and  the  point  of 
sight  be  a  great  distance  from  the  object,  all  the  rays  will  make 


§3 


REPRESENTING  SOLIDS  ON  PLANES 


3 


angles  with  the  picture  plane  that  are  very  nearly  equal  to  a 
right  angle;  and,  if  the  point  of  sight  be  assumed  to  be  situated 
at  an  infinite  distance  from  the  object,  the  rays  will  all  be  parallel 
and  will  all  be  perpendicular  to  the  picture  plane.  The  resulting 
drawing  will  then  be  a  projection  of  the  object  on  the  picture 
plane,  as  shown  in  Fig.  2.  Here  only  one  side  of  the  object  is 
represented  on  the  drawing,  the  side  CBGF;  the  entire  line  AB  is 
projected  in  the  single  point  a,  6;  the  lines  DC,  EF,  and  the  line 
representing  the  edge  running  back  from  G  are  also  projected  in 
single  points,  since  they  are  all  perpendicular  to  the  picture  plane. 
It  will  be  noted  that  the  hole  does  not  appear  at  all. 

3.  While  the  drawing  in  Fig.  2  shows  only  one  side  of  the  object 
and  gives  scarcely  any  intima- 
tion as  to  the  shape  of  the  object, 
it  nevertheless  possesses  several 
advantages  over  that  shown  in 
Fig.  1.  On  the  frustum,  the  lines 
AB  and  CD,  also  AD  and  CB, 
are  equal  and  parallel,  but  in 
Fig.  1,  these  lines  are  neither 
equal  nor  parallel,  and  it  would 
be  very  difficult  to  determine 
from  the  lengths  of  the  lines  ab, 
be,  etc.  the  true  lengths  of  the 
edges  AB,  BC,  etc.  In  Fig.  2, 
cb  and  fg  are  parallel,  as  they 
should  be;  cb  =  CB  and  fg  - 
FG.  Further,  by  using  dotted  lines  to  indicate  lines  that  are 
hidden,  the  depth  and  diameter  of  the  hole  can  also  be  shown. 

4.  Projection  Drawings. — The  drawing  shown  in  Fig.  1  is 
called  a  scenographic  projection  drawing,  a  perspective  drawing, 
or,  simply,  a  perspective ;  the  drawing  shown  in  Pig.  2  is  called  an 
orthographic  projection  drawing  or,  simply,  a  projection  drawing. 
In  the  majority  of  cases  that  arise  in  practice,  two  views  are  given 
of  the  object,  one  view  being  taken  parallel  to  the  horizontal 
plane  and  the  other  parallel  to  the  vertical  plane;  and  when  both 
these  views  are  projection  drawings  and  are  fully  dimensioned, 
they  will  usually  suffice  for  reproducing  the  object  in  its  exact 
size.  If  the  object  is  a  complicated  machine  or  machine  part, 
three  views  may  be  necessary.     If,  in  addition,  the  object  has  a 


Fig. 


4  HOW  TO  READ  DRAWINGS  §3 

complicated  interior  arrangement  that  is  hidden  by  the  outside 
surface  of  the  object,  interior  views  (sections)  are  frequently- 
necessary  or  desirable,  in  order  to  obviate  the  use  of  too  many 
lines,  which  may  make  the  drawing  very  hard  to  read.  These 
features  will  be  made  clear  in  what  follows. 

Suppose  the  frustum  of  Fig.  1  be  enclosed  in  a  glass  box,  as 
illustrated  in  Fig.  3,  and  that  the  front  and  right-hand  sides  and 
the  top  be  treated  as  picture  planes,  the  glass  being  assumed  to  be 
transparent.  Projecting  the  object  on  the  front  plane  M,  the  result 
is  the  outline  d"c"f'e",  which  may  be  regarded  as  a  drawing  of 
the  frustum  on  the  picture  plane  M  when  the  eye  is  at  an  infinite 
distance  from  it.  The  lines  Dd" ,  Cc",  etc.,  are  the  light  rays 
from  the  object  to  the  eye;  as  applied  to  drawings,  these  lines  are 
called  projectors.  In  a  similar  manner,  Cc'",  Bb'",  etc.  are 
projectors  from  the  object  to  the  plane  side  P,  and  c"'b"'g'"f'" 
is  the  projection  of  the  frustum  on  the  plane  P,  considered  as  a 
picture  plane.  Likewise,  Ee',  Dd',  etc.  are  projectors  from  the 
object  to  the  top  plane  N,  and  the  outline  there  shown  is  a  pro- 
jection drawing  on  the  plane  N,  considered  as  a  picture  plane. 
Note  that  the  hole  is  shown  in  all  three  views,  being  represented 
as  a  circle  in  the  top  plane,  and  as  a  dotted  rectangle  in  the  front 
and  side  planes. 

The  line  of  intersection  made  on  a  plane  A  by  the  intersection 
with  it  of  another  plane  B  is  called  the  trace  of  the  plane  B; 
Thus,  in  Fig.  3,  the  intersection  of  plane  M  with  plane  A7"  is  the 
line  JK,  and  JK  is  the  trace  of  plane  M  on  plane  N;  likewise, 
JK  is  the  trace  of  plane  N  on  plane  M.  Similarly,  KL  is  the 
trace  of  plane  P  on  plane  N,  and  KI  is  the  trace  of  plane  P  on 
plane  M.  The  lines  JK,  KL,  and  KI  are  so  important  that 
they  have  received  special  names :  JK  is  called  the  front  trace ; 
KL  is  called  the  side  trace;  and  KI  is  called  the  vertical  side 
trace. 

If  a  plane  be  passed  through  the  line  EF  (the  lower  front  edge 
of  the  frustum)  parallel  to  the  plane  M,  the  trace  of  this  plane 
on  N  will  be  the  line  e'fiv,  and  the  trace  on  plane  P  will  be  the 
line  fivf".  (A  plane  is  of  infinite  extent.)  It  will  be  observed 
that  e'fiv  is  parallel  to  the  projector  Ff";  in  fact,  it  is  a  projec- 
tor from  /'  to  the  side  trace,  which  it  intersects  in  fiv.  Also, 
fivf"  is  parallel  to  the  projector  Ee'  and  is  a  projector  from/'" 
to  the  side  trace  KL,  which  it  intersects  in  fiv  also.  Similarly, 
ev  is  the  projector  of  e'  and  e"  on  the  front  trace  JK,  and  is  the 


§3 


REPRESENTING  SOLIDS  ON  PLANES 


point  of  intersection  of  the  traces  e"ev  and  evh'  of  the  plane 
h'eve"H,  which  has  been  passed  through  EH  parallel  to  plane 
P.  Consequently,  if  the  position  of  the  projection  of  any  point 
on  the  planes  JV  and  M,  N  and  P,  or  M  and  P  is  known,  the 
position  of  the  projection  of  the  point  on  the  third  plane  can 
easily  be  found.  For  example,  suppose  the  positions  of  c'  and  c" 
are  known,  and  it  is  desired  to  find  the  position  of  this  point 
on  plane  P.  Draw  c'civ  perpendicular  to  KL,  which  it  intersects 
in    civ;    through    civ,    draw    civc'"    perpendicular    to    KL;    also 


Fig.  3. 

c"cvi  perpendicular  to  A'7,  which  it  intersects  in  cvi;  draw 
cvic'"  perpendicular  to  KI  (and  parallel  to  KL) ;  cvic'"  intersects 
civc'"  in  c"',  which  is  the  position  of  the  projection  of  point  C 
of  the  object  on  the  side  plane  P.  If  the  projections  c"  and  c'" 
had  been  given  and  the  projection  c'  on  the  plane  N  had  been 
desired,  project  c"  on  the  front  trace  in  cv;  project  c'"  on  the 
side  trace  in  civ;  draw  cvc'  and  c'V  perpendicular  respectively 
to  JK  and  KL,  and  they  intersect  in  c',  which  is  the  required 
projection. 

_  The  outlines  a'b'c'd'e'f'gV,  d"c"f"c",  and  c"b'"q"'S"'  are  projec- 
tion drawings  of  the  rectangular  frustum;  they  arc  not  suitable  as 
they  stand  for  use  as  working  drawings  or  for  any  other  purpose. 
To  make  them  suitable,  imagine  the  side  M  of  the  glass  box  to 
be  hinged  to  the  topside  N,  and  the  side  P  to  be  hinged  to  the  top 


6  HOW  TO  READ  DRAWINGS  §3 

also.  Now  lift  M  until  its  plane  coincides  with  the  plane  of  N,  and 
lift  P  until  its  plane  coincides  with  the  plane  of  N  also;  the  result 
will  be  as  shown  in  Fig.  4.  Note  that  the  projectors  g'giv  and 
girg'",  b'biv  and  bivb'",  etc.  become  a  single  right  line  in  Fig. 
4;  and  the  same  thing  is  true  of  the  projectors  e'ev  and  cve" , 
d'dv  and  dvd",  etc.  In  other  words,  g'  may  be  projected 
directly  to  g'",  b'  to  b'",  etc.;  and  e'  may  be  projected  directly 
to  c',  d'  to  d",  etc. ,  the  only  requisite  being  that  the  distance  between 
g'"f"  and  b'"c'"  or  the  distance  between  e"f"  and  d"c"  must 
be  known,  and  this  will  be  given  by  measurements  made  on  the 
object. 

Instead  of  the  arrangement  of  views  shown  in  Fig.  4,  another 
arrangement  equally  correct  may  be  and  is  frequently  employed. 
Referring  to  Fig.  3,  suppose  that  instead  of  hinging  the  side  P  to 
the  top  N,  the  side  P  is  hinged  to  the  front  M;  then  pulling  out  the 
side  P  until  its  plane  coincides  with  the  plane  of  M ,  lift  both 
planes  until  the  common  plane  of  M  and  P  coincides  with  the 
plane  of  N.  The  result  will  be  the  arrangement  shown  in  Fig.  5. 
The  arrangement  shown  in  Fig.  4  is  the  more  natural,  but  that 
shown  in  Fig.  5  may  sometimes  be  more  convenient.  Either 
Fig.  4  or  Fig.  5  is  a  correct  projection  drawing  of  the  frustrum, 
and  when  properly  dimensioned,  may  be  used  as  a  working 
drawing. 

5.  Names  of,  and  Number  of,  Views. — Referring  to  Figs.  4 
and  5,  view  JV  is  called  either  a  top  view  or  a  plan;  it  is  the  view 
obtained  when  the  eye  is  directly  over  the  object.  View  M  is 
called  either  a  front  view  or  a  front  elevation  or,  simply,  the  eleva- 
tion ;  it  is  the  view  obtained  when  the  eye  is  directly  in  front  of  the 
object.  View  P  is  called  a  side  view  or  a  side  elevation;  it  is  the 
view  obtained  when  the  eye  is  so  situated  as  to  see  the  object 
from  the  side,  the  picture  plane  being  at  right  angles  to  the  front 
and  top  planes. 

In  both  Figs.  4  and  5,  three  views  are  given,  and,  theoretically, 
three  views  on  planes  perpendicular  to  one  another  are  sufficient 
to  give  the  size  and  shape  of  any  object,  since  no  solid  has  more 
than  three  dimensions — length,  breadth,  and  thickness.  In 
practice,  however,  it  is  sometimes  advisable  to  show  more  than 
three  views,  because  the  number  of  lines  on  the  drawing  would 
otherwise  be  so  numerous  that  the  drawing  would  be  very  difficult 
to  read.  Since  the  glass  box  in  Fig.  3  has  six  sides,  it  is  possible 
to  get  six  views  without  modifying  the  shape  of  the  box.     Thus 


§3 


REPRESENTING  SOLIDS  ON  PLANES 


what  is  termed  a  bottom  view  or  inverted  plan  is  obtained  by 
using  TQRI  as  a  picture  plane,  another  side  view  by  using  TQSJ 
as  a  picture  plane,  and  a  back  view  or  rear  elevation  by  using 
RQSL  as  a  picture  plane.  To  get  these  views  into  proper  posi- 
tion, imagine  the  bottom  plane  to  be  hinged  to  plane  M,  the  side 
planes  to  be  hinged  to  plane  N,  and  the  back  plane  to  be  hinged  to 
plane  N  also.     Now  revolve  the  bottom  plane  TQRI  downward 


h' 

9 

P 

ff" 

r 
,1 

Y 

i\ 

/ 

ff'v 

\. ■ " — 

4 

!  • 



b" 

1 

^^^^^^ 

1 

^ 

\     . 

0"        c 

el 

1  » 

c  I 

V 

f" 

1 

Id 

i 

1 
C 

f 

K 

I 

• 

1 
M       1 

J.J,                  1 

c 

J 

Fig.  4. 

until  it  coincides  with  plane  M,  and  then  revolve  both  planes  up- 
ward until  they  coincide  with  plane  X,  likewise,  revolve  planes 
TQSJ,  IRLK,  and  RQSL  upward  until  they  also  coincide  with 
plane  N;  the  result  will  be  as  shown  in  Fig.  6.  Except  for  the 
letters  at  the  corners,  the  side  views  are  alike  and  the  front  and 
back  views  are  also  alike,  since  the  object  is  symmetrical;  the 
bottom  view,  however,  is  somewhat  different  from  the  top  view 
by  reason  of  the  dotted  lines,  which  are  used  because  the  base  of 
the  frustum  hides  everything  above  it  when  tin1  frustum  is  viewed 
from  below. 


8 


HOW  TO  READ  DRAWINGS 


§3 


When  all  six  views  are  shown,  the  arrangement  in  Fig.  6  is  to  be 
preferred.  However,  the  bottom  and  sides  can  be  imagined  to  be 
hinged  in  any  other  manner  desired  that  will  permit  all  the  planes 
to  be  brought  into  coincidence  with  the  top  plane.  Conse- 
quently, the  inverted  plan  may  be  placed  to  the  right  of  the 
right  side  elevation  c"'h'"g'"j'" ,  to  the  left  of  the  left  side  eleva- 
tion dviiaviihviievii,  or  above  the  rear  elevation  avibvigvihvi,  and  the 


h 

• 

ff' 

\ 

u_ 

i 

1 

O 

J 

d     | 

c 

\; 

\        \ 

\ 

e~ 

d 

1/ 
c 

K 

£ 

e1 

d"   \  i 

c 

rv 

c 

i" 

h" 

1 

\     P 

M 

l£ 

t 

J". 

\ 

e 

f 

/'" 

\  m 
9 

R 

Fig.  5. 

two  side  elevations  may  be  placed  to  the  right  and  left  of  the 
front  elevation,  as  noted  in  Fig.  5,  if  desired.  It  will  be  noted 
that  the  letters  at  the  outside  corners  on  the  inverted  plan  are 
capital  letters,  corresponding  to  those  at  the  corners  of  the  base 
of  the  frustum,  because  as  shown  in  Fig.  3  the  frustum  is  sup- 
posed to  rest  on  the  bottom  plane  and  there  are,  consequently, 
no  projectors  from  the  base  to  the  picture  plane. 

6.  Working  Drawings. — A  working  drawing  is  a  projection 
drawing  on  which  all  necessary  dimensions  are  marked  and  on 
which  all  necessary  notes  are  written  or  printed  that  are  required 


§3  REPRESENTING  SOLIDS  ON  PLANES 


/ 


b" 


11 


Mill  Vw 


f 


/' 


Fiu.   6. 


10 


HOW  TO  READ  DRAWINGS 


§3 


in  order  that  the  object  represented  may  be  made  or  reproduced. 
A  working  drawing  of  the  frustum  of  Fig.  1  is  shown  in  Fig.  7. 
It  will  be  observed  that  only  two  views  are  given;  two  views  are 
all  that  are  required  in  this  case,  as  the  following  considerations 
will  show: 

The  lines  ?nn  and  pq  are  called  center  lines;  in  the  present  case, 
they  are  axes  of  symmetry  in  the  plan  and  mn  is  an  axis  of  sym- 
metry in  the  front  elevation.  It  is  plainly  evident  that  the  point 
of  intersection  of  mn  and  pq  is  the  center  of  the  circle  that  repre- 
sents the  plan  of  the  hole.     The  plan  shows  that  there  are  two 


i 


♦-j-jt  \-'*'-jf&"    -K^f-    ,j~g*!~i"/,<^'" 


-4i- 


W 


Fig.  7. 


Fig.  8. 


surfaces,  ABCD  and  EFGH,  whose  projections  are  rectangles,  and 
the  elevation  shows  that  these  surfaces  are  planes  (flat  surfaces) 
and  that  they  are  parallel.  If  they  were  not  plane  surfaces,  the 
lines  DC  and  EF  in  Fig.  7  would  not  be  straight,  and  if  they  were 
not  parallel,  the  lines  DC  and  EF  would  not  be  parallel.  It  is 
seen  from  the  dimensions  that  EF  and  HG  are  both  4  in.  from  the 
center  line  pq,  and  that  EH  and  FG  are  both  2|  in.  from  the 
center  line  mn;  also,  A B  and  DC  are  both  2f  in.  from  pq,  and 
DA  and  CB  are  both  if}  in.  from  mn.  Since  these  four  dimen- 
sions are  given  only  once,  it  is  inferred  that  HG,  AB,  DC,  and  EF 
are  parallel  to  pq,  and  that  EH,  DA,  CB,  and  FG  are  parallel  to 


§3  REPRESENTING  SOLIDS  ON  PLANES  11 

mn,  thus  making  HGFE  and  ABCD  rectangles,  since  the  main 
center  lines  are  always  drawn  perpendicular  to  each  other.  The 
two  planes  are  parallel  and  are  located  symmetrically  with 
respect  to  the  center  lines  mn  and  pq;  their  distance  apart  is 
given  in  the  elevation,  and  is  found  to  be  6|  in.  The  general 
shape  is,  consequently,  a  prismoid,  since  if  the  sides  were  not 
plane  surfaces,  some  of  the  lines  connecting  the  top  and  bottom, 
as  DE,  CF,  BG,  and  AH  (in  both  plan  and  elevation)  would  not  be 
straight.  The  hole  is  11  in.  in  diameter  and  3  in.  deep,  and 
extends  down  from  the  top.' 

To  make  the  object  from  say,  a  piece  of  wood,  a  rectangular 
prism  having  a  cross  section  of  4§"  X  8"  and  a  length  of  6\" 
would  be  made.    On  one  end,  lines  would  be  drawn  parallel  to 
the  long  sides  and  2\  -  lf£  =  &  in.  from  them;  also,  lines 
would  be  drawn  parallel  to  the  short  sides  and  4  -  2f  =  If  in. 
from  them,   as  indicated  in  the  elevation,   Fig.  8.     Then  the 
material  included  between  the  outer  edges  and  the  dotted  lines 
would  be  planed  off,  the  result  being  the  frustum.     The  center  of 
the  hole  could  then  be  located  by  drawing  the  diagonals,  as  indi- 
cated in  the  plan,  or  by  laying  off  ab  and  cd,  as  indicated,  both 
being  equal  to  lf£",  and  drawing  bd;  then  lay  off  bo  or  do  equal 
to  2f",  thus  locating  the  center  o.     Having  found  the  position  of 
o,  drill  or  bore  a  hole  l\"  in  diameter  and  3"  deep,  the  axis  of  the 
hole  being  perpendicular  to  the  bases,  and  the  work  is  completed. 
It  will  be  noticed  that  no  projectors  are  shown  in  Figs.  7  or  8; 
they  are  never  shown  on  finished  drawings  or  on  working  draw- 
ings.    When  reading  a  drawing,  that  is,  when  studying  it  to  find 
out  the  shape  of  the  object  and  the  details  of  its  parts,  the  pro- 
jectors can  always  be  imagined  to  be  present;  and  if  the  eye  does 
not  readily  perceive  the  connection  between  different  parts  of 
two  views,  this  may  usually  be  found  by  means  of  a  straightedge 
assisted,  perhaps,  with  dividers  to  set  off  distances.     This  feature 
will  be  dwelt  on  more  fully  later. 


SPECIAL  FEATURES  PERTAINING  TO  DRAWINGS 

7.  Different  Kinds  of  Lines  Used  on  Drawings. — In  Fig.  9,  are 
shown  five  different  lines,  which  are  used  in  the  following  manner: 

Line  I. — This  is  a  full  line  and  may  be  of  any  convenient 
weight  (by  weight  is  here  meant  thickness).  This  line  should 
have  the  same  weight  wherever  used  on  any  particular  drawing; 


12  HOW  TO  READ  DRAWINGS  §3 

it  is  employed  in  all  cases  when  the  outline  of  the  object  can  be 
seen  with  the  eye  in  the  position  it  is  assumed  to  occupy  when 
the  view  is  drawn.  This  line  is  used  more  than  any  of  the  others 
and  is.  consequently,  the  most  important. 

Line  II. — Tins  line,  called  the  dotted  line,  consists  of  a  succes- 
sion of  dots  or  very  short  dashes;  it  is  used  to  show  the  outline  of 
parts  that  cannot  be  seen  by  the  eye  when  in  its  assumed  position 
relative  to  the  view  being  drawn.  The  inverted  plan,  or  bottom 
view,  in  Fig.  6  is  a  good  example  of  the  use  of  the  dotted  line. 
Viewing  the  frustum  from  the  bottom  all  that  can  be  seen  is  the 
outline  of  the  base,  which  is  drawn  with  full  lines.  The  top  and 
the  edges  connecting  the  top  and  bottom  are  then  represented  by 
dotted  lines,  and  likewise  the  hole.  This  line  is  never  used  for 
any  purpose  other  than  to  represent  invisible  outlines. 

I 

I 

m 

w 

Y 


Fig.   9. 

Line  III. — The  broken  and  dotted  line  consists  of  a  series  of 
long  dashes  with  a  single  dot  or  very  short  dash  between  the  long 
dashes;  it  is  used  for  center  lines,  and  is  also  frequently  employed 
to  denote  the  traces  of  planes,  showing  where  a  section  has  been 
taken.  Its  use  as  a  center  line  is  shown  in  Figs.  6  and  7.  Its  use 
to  indicate  where  a  section  is  taken  will  be   discussed  later. 

Line  IV. — This  broken  and  dotted  line  consists  of  a  succession 
of  long  dashes  and  two  dots;  it  is  used  for  the  same  purpose  as 
line  III.  and  both  rarely  appear  on  the  same  drawing,  but  when 
they  do,  line  III  is  used  for  center  lines  and  line  IV  to  indicate 
where  a  section  has  been  taken. 

Line  V. — The  broken  line  consists  of  a  series  of  long  dashes;  it 
is  generally  used  for  the  same  purpose  as  projectors,  to  indicate 
the  extension  or  prolongation  of  lines,  but  its  principal  use  is  for 
dimension  lines.  It  is  also  sometimes  used  to  indicate  the 
extension  of  lines  that  are  not  actually  a  part  of  the  view  being 
drawn.  Fig.  7  shows  how  it  is  employed  in  extending  lines  and 
in  dimension  lines. 

While  the  foregoing  specifies  the  manner  in  which  these  lines 


§3  REPRESENTING  SOLIDS  ON  PLANES  13 

are  most  generally  used,  there  are  occasional  exceptions,  particu- 
larly in  the  case  of  working  drawings.  In  some  drafting  offices, 
the  regular  full  line  is  made  very  heavy  and  a  light  full  line 
is  used  in  the  same  manner  as  line  III;  this  practice,  however,  is 
not  recommended.  Again,  it  is  quite  common  practice  to  use 
the  light  full  line  for  dimension  lines,  it  being  broken  only  where 
the  dimension  is  placed.  But,  when  lines  III,  IV,  and  V  are 
used,  they  are  usually  employed  as  specified  above. 

8.  Center  Lines. — Center  lines  are  used  for  two  purposes:  for 
what  may  be  termed  base  lines,  from  which  to  take  measurements; 
and  to  indicate  curved  surfaces,  particularly  circles  and  cylindrical 
surfaces.     They  are  also  used  as  axes  of  symmetry.     Referring 
to  Fig.  6,  the  center  line  pq  serves  two  purposes:  it  is  an  axis  of 
symmetry  for  the  plan  and  the  two  side  elevations;  it  shows, 
in  connection  with  the  circle  in  the  plan,  that  the  hole  is  round— 
a  right  circular  cylinder,  in  fact.     The  center  line  mn  also  serves 
the  same  two  purposes,  being  an  axis  of  symmetry  for  the  plan, 
the  front  and  rear  elevations,  and  the  inverted  plan.     The  point 
of  intersection  of  these  two  center  lines  locates  the  center  of  the 
circle  that  represents  the  top  and  bottom  views  of  the  hole.     On 
working  drawings,  center  lines  are  invariably  drawn  to  define  the 
axis  of  cylindrical  surfaces,  and  their  presence  on  a  drawing 
passing  through  a  rectangle  is  sufficient  in  itself,  as  a  rule,  to  show 
that  the  rectangle  is  the  projection  of  a  cylinder.    In  the  case  of  a 
circle,  two  center  lines  are  always  drawn  at  right  angles  to  each 
other,  thus  locating  the  center  of  the  circle.     The  use  of  center 
lines  as  base  lines  is  illustrated  in  Fig.  7;  here  the  dimensions 
2?",  2\" ,  21",  2f  "  etc.  which  are  measured  from  the  center  lines 
outward,  serve  to  locate  the  position  of  the  hole,  and  they  also 
show  by  their  equality  that  the  center  lines  are  axes  of  symmetry. 

9.  Dimension  Lines.— Whenever  it  is  possible  to  avoid  it, 
dimension  lines  are  seldom  drawn  across  the  face  of  any  view; 
this  is  to  prevent  confusion  in  reading  the  drawing,  and  also  to 
make  the  dimensions  more  prominent.  By  the  use  of  extension 
lines,  the  dimension  lines  and  dimensions  arc  very  largely  kept 
off  the  different  views.  Practically  the  only  exception  to  this 
rule  is  when  giving  the  diameters  of  circles  and  the  radii  of  cir- 
cular arcs.  The  use  of  extension  lines  is  shown  in  Figs.  6  and  7. 
In  Fig.  7  it  will  be  noted  that  the  diameter  of  the  hole  is  printed 
on  the  front  elevation,  thus  indicating  the  diameter  of  the  cylin- 


14  HOW  TO  READ  DRAWINGS  §3 

drical  hole  instead  of  the  diameter  of  the  circle  in  the  plan;  this  is 
done  because  of  the  crossing  of  the  center  lines  on  the  circle. 

When  there  is  room  enough,  the  dimension  lines  have  arrow- 
heads, one  at  each  end,  with  the  dimension  written  or  printed 
about  midway  between  the  ends;  but,  when  the  space  is  limited, 
as  in  the  case  of  the  diameter  of  the  hole  in  Fig.  7,  short  arrows 
with  their  heads  pointing  toward  each  other  are  used,  the  dimen- 
sion being  placed  between  the  arrowheads.  In  some  cases, 
even  this  will  not  suffice,  the  dimension  being  placed  alongside 
the  arrow,  as  illustrated  in  Fig.  8  in  connection  with  the  dimen- 
sion -jV". 

What  are  termed  the  general  or  over  all  dimensions  are  those 
relating  to  the  extreme  or  outer  bounding  lines  of  the  figure.  In 
Fig.  7,  the  over  all  dimensions  are  8",  4|",  and  6£",  the  first 
two  giving  the  size  of  the  base  and  the  third  the  height  of  the 
frustum.  The  general  dimensions  include  these  and  also 
b\"  and  2>\\",  which  give  the  size  of  the  top. 

It  might  be  argued  that  both  of  the  two  dimensions,  which 
added  together  make  the  over  all  dimension  8"  are  not  necessary, 
since  if  one  is  given  the  other  can  be  found  by  subtracting  from 
8;  this  is  true,  but  the  idea  of  giving  both  is  to  save  subtracting, 
which  is  sometimes  awkward.  For  instance,  the  distance 
AB,  Fig.  7,  is  31!";  if  the  distance  from  A  to  the  center  line 
were  given  as  lfl",  it  is  not  readily  apparent  that  the  dis- 
tance from  B  to  the  center  line  is  if  i",  and  this  can  be  found 
only  by  subtracting  lfi"  from  3t!",  a  somewhat  awkward 
operation  and  one  that  takes  a  certain  amount  of  time,  together 
with  the  possibility  of  making  a  mistake.  It  is  for  the  same  reason 
that  the  over  all  dimensions  are  given — to  obviate  the  necessity 
of  adding  the  intermediate  dimensions.  Over  all  dimensions 
are  placed  outside  of,  that  is,  beyond,  all  shorter  measurements. 

10.  Scales. — Up  to  this  point,  drawings  have  been  described  as 
though  every  line  on  them  was  of  the  same  length  as  the  corre- 
sponding measurements  taken  on  the  object.  It  is  obvious  that 
this  is  not  feasible  in  many  cases,  for  not  only  would  it  be  extremely 
difficult  to  make  the  drawing  but  it  would  also  be  very  difficult  if 
not  impossible  to  read  it.  The  matter  of  storing  and  preserving 
drawings  must  also  be  considered.  Consequently,  most  of  the 
drawings  in  actual  use  are  made  to  scale,  as  it  is  termed;  by 
this  is  meant  that  the  drawing  is  smaller  than  if  all  the  dimensions 
corresponded  in  length  with  those  of  the  object.     In  such  cases, 


§3  REPRESENTING  SOLIDS  ON  PLANES  15 

lines  on  the  drawing  arc  only  one-half,  one-third,  one-fourth,  etc. 
as  long  as  those  they  represent  on  the  object.  Sometimes, 
when  the  object  is  small  and  it  is  desired  to  bring  out  certain 
details  prominently,  the  drawing  may  be  made  two,  three,  four, 
etc.,  times  as  large  as  the  object. 

Any  drawing  that  has  been  made  accurately,  every  line  being 
carefully  measured  with  a  scale,  is  called  a  scale  drawing.  If 
every  line  on  the  drawing  is  of  the  same  length  as  the  line  it 
represents  on  the  object,  the  scale  is  full  size  or  12"  =  1  ft.  If  a 
line  on  the  drawing  is  only  half  as  long  as  the  corresponding  line 
on  the  object,  the  scale  is  half  size  or  6"  =  1  ft.  if  the  lines  on  the 
drawing  are  only  a  quarter,  an  eighth,  etc.  as  long  as  the  corre- 
sponding lines  on  the  object,  the  scale  is  quarter  size,  eighth  size 
etc.  A  quarter-size  scale  is  3"  =  1  ft.  because  one-fourth  of 
12"  is  3";  hence,  3"  measured  on  the  drawing  equal  1  ft.  measured 
on  the  object.  If  3"  be  divided  into  12  equal  parts,  each  part 
will  represent  1  inch  on  the  object  when  the  drawing  is  made  to  a 
scale  of  3"  =  1  ft.  Dividing  each  of  these  parts  into  halves, 
quarters,  eighths,  etc.,  these  smaller  divisions  represent  halves, 
quarters,  eighths,  etc.  of  an  inch. 

The  scales  most  commonly  used  are:  full  size  (12"  =  1  ft.), 
half  size  (6"  =  1  ft.),  3"  =  1  ft.  (quarter  size),  1|"  =  1  ft, 
(eighth  size),  and  f"  =  1  ft.  (sixteenth  size).  The  first  three 
scales  are  the  most  used,  and  a  scale  smaller  than  f"  =  1  ft. 
is  very  seldom  needed  in  drawings  of  machines.  Architects  and 
civil  engineers  use  scales  very  much  smaller,  a  favorite  scale 
with  architects  being  1"  =  4  ft.  (or  \"  =  1  ft.). 

Any  drawing  that  has  been  made  to  scale  should  always  have 
the  scale  used  marked  on  it.  It  sometimes  happens  that  a  part 
of  the  drawing  may  be  made  to  one  scale  and  a  part  to  another 
scale;  in  eveiy  such  case,  the  scale  used  should  always  appear  on 
the  drawing  in  connection  with  the  part  to  which  it  applies. 

To  use  a  scale  to  find  the  actual  length  of  a  particular  line  on 
the  object,  proceed  as  follows:  suppose  the  scale  is  3"  =  1  ft,, 
and  that  an  ordinary  12-inch  scale,  with  inches  divided  into 
halves,  quarters,  eighths,  sixteenths,  and  thirty-seconds,  is  used. 
Suppose  further  that  the  actual  length  of  the  line  on  the  drawing 
that  is  being  measured  is  33f  in.  Since  the  scale  is  3"  =  1  ft., 
1  in.  on  the  drawing  represents  12  -4-  3  =  4  in.  on  the  object; 
j  in.  on  the  drawing  represents  1  in.  on  the  object;  and  3V  in. 
on  the  drawing  represents  -3V  X  4  =  f  in.  on  the  object,     Con- 


16  HOW  TO  READ  DRAWINGS  §3 

sequently,  3  jf  in.  on  the  drawing  represent  3X4  +  13X|  = 
12  +  If  =  13f  in.  on  the  object.  Had  the  scale  been  \\"  =1 
ft.,  1  in.  on  the  drawing  =  12  -5-  If  =  8  in.  on  the  object;  yV 
in.  on  the  drawing  =yjX8=  jin.  on  the  object;  and  3ff  in. 
on  the  drawing  represent  3  X8+13Xi  =  24+  31  =  27  \  in. 
on  the  object.  Observe  that  since  the  second  scale  is  twice  as 
small  as  the  first,  the  length  of  the  line  when  measured  to  the 
second  scale  should  be  twice  as  long  as  when  measured  to  the 
first  scale;  and  this  is  the  case,  since  13f  X  2  =  27 j. 

The  process  of  determining  the  length  of  aline  on  the  object  by 
measuring  the  corresponding  line  on  the  drawing  is  called  scaling 
the  drawing,  and  when  a  measurement  has  thus  been  taken,  the 
drawing  is  said  to  be  scaled.  Any  drawing  may  be  scaled  pro- 
vided the  scale  to  which  the  drawing  was  made  is  known,  pro- 
vided further  that  the  drawing  has  been  made  accurately  and  the 
scale  used  for  measuring  is  accurate.  Special  rules  (scales)  are 
made  for  making  drawings  to  scale  and  measuring  them. 

11.  Abbreviations  and  Notes  on  Drawings. — When  a  line  is 
drawn  from  the  center  to  the  arc  to  indicate  a  radius,  the  abbre- 
viation r.  or  rad.  is  almost  invariably  written  after  the  dimension, 
thus  making  it  clear  that  the  dimension  is  the  radius  of  the  arc. 
An  arrowhead  is  placed  at  the  end  of  the  dimension  line  touching 
the  arc,  but  not  at  the  center.  The  abbreviation  D.,  Dia., 
Diam.,  d.,  dia.,  or  diam.  is  used  to  follow  the  dimension  indicating 
the  diameter  of  a  circle  or  a  cylindrical  surface.  Thds  or  thds 
means  threads;  thus,  8  thds  means  8  threads  to  the  inch,  and  refers 
to  screw  threads.  The  letter/,  or  fin.  means  finish,  and  indicates 
that  the  surface  on  which  it  is  written  in  the  drawing  is  to  be 
finished.  If  only  a  part  of  the  surface  is  to  be  finished,  this  is 
frequently  indicated  by  drawing  a  line  near  to  and  parallel  to  the 
line  representing  the  surface,  but  extending  only  as  far  as  the  sur- 
face is  to  be  finished,  and  writing  on  it  /.  or  fin. 

In  addition  to  the  dimensions,  working  drawings  usually  carry  a 
number  of  "notes,"  which  are  written  or  printed  on  the  drawings 
for  the  benefit  of  the  workmen.  In  connection  with  these  notes, 
certain  terms  are.  employed  that  deserve  explanation.  Thus,  the 
word  cored  means  that  the  hole  is  to  be  made  by  means  of  a  core 
when  casting  and  is  not  to  be  finished;  bore  or  bored  means  that 
the  hole  is  to  be  cored  and  finished  by  boring;  drill  means  that  the 
hole  is  to  be  made  by  drilling ;  ream  or  reamed  means  that  after  the 
hole  has  been  bored  or  drilled  is  to  be  finished  by  reaming;  the 


§3  REPRESENTING  SOLIDS  ON  PLANES  17 

word  tap  means  that  the  hole  is  to  be  threaded,  a  tap  being 
used  for  this  purpose;  faced  means  that  the  surface  is  to  be 
finished  in  a  lathe  or  boring  mill,  or  other  machine  tool  that  will 
make  the  surface  flat — usually  by  revolving  it  against  the  cutting 
tool;  planed  means  that  the  surface  is  to  be  finished  by  planing, 
the  tool  used  being  a  planer,  shaper,  or  a  milling  machine;  grind 
means  that  the  surface  is  to  be  finished  by  grinding;  scraped 
means  that  the  surface  is  to  be  finished  by  scraping,  that  is, 
hand  finished  with  a  scraper;  tool  finish  means  that  after  the  sur- 
face has  been  finished  on  the  machine,  nothing  further  is  to  be 
done  to  it.  Other  terms  are  sometimes  used,  but  they  are  usually 
self  evident  and  need  not  be  referred  to  here. 


SECTIONS  AND   SECTIONAL  VIEWS 

12.  Why  Sections  are  Used. — While  all  hidden  surfaces  and 
parts  may  be  represented  on  a  drawing  by  dotted  lines,  it  is 
nevertheless  frequently  advisable  to  show  what  are  called  sec- 
tional views  or  sections  instead  of  making  the  drawing  in  the 
regular  manner.  The  cutting  plane  may  be  considered  to  pass 
through  every  part  of  the  object  that  it  can  touch  or  its  trace  may 
be  limited  to  only  a  short  length  relating  merely  to  a  single 
detail.  Sections  are  usually  so  taken  that  lines  which  would 
appear  dotted  in  a  regular  projection  drawing  appear  as  full  lines 
in  the  sectional  view,  thus  making  the  drawing  easier  to  read. 
This  fact  is  brought  out  in  Fig.  10,  which  is  a  drawing  of  a  clamp- 
ing ring.  The  center  lines  mn  and  pq  are  axes  of  symmetry. 
The  view  in  the  middle  may  here  be  considered  as  a  front  eleva- 
tion, the  view  on  the  right  being  a  side  elevation  and  that  on  the 
left  a  sectional  elevation.  Note  how  much  clearer  the  sectional 
elevation  is  than  the  side  elevation. 

Considering  the  sectional  view,  the  section  is  taken  along  the 
center  line  mn,  the  cutting  plane  being  perpendicular  to  the  flat 
surface  included  between  the  circles  c  and  g,  that  is,  perpendicular 
to  the  plane  of  the  paper;  that  part  of  the  ring  to  the  left  of  mn  is 
then  imagined  to  be  removed,  and  a  drawing  is  made  of  the  re- 
maining part,  the  eye  being  situated  to  the  left  of  mn.  Now  note 
particularly  that  whenever  a  section  is  taken,  any  surface  touched 
by  the  cutting  plane  is  indicated  on  the  drawing  by  cross  hatch- 
ing, as  it  is  termed,  the  cross  hatching  consisting  of  parallel  lines, 
drawn  at  an  angle,  usually  45°,  to  the  horizontal.      By  using 


IS  HOW  TO  READ  DRAWINGS  §3 

different  kinds  of  cross  hatching,  different  materials  may  be  in- 
dicated, bo  that  an  inspection  of  the  section  lines  (cross  hatch- 
ing) will  show  at  once  what  material  is  to  be  used  in  making  the 
object  represented  by  the  drawing.  Thus,  in  Fig.  10,  the  cross 
hatching  there  used  represents  steel,  as  will  be  explained  pres- 
ently ;  hence,  the  clamping  ring  is  to  be  made  of  steel.  Supposing 
the  right-hand  view  to  be  omitted,  the  drawing  may  then  be  read 
as  follows : 

All  dimension  lines  extending  up  and  down  and  parallel  to  the 
center  line  mn  are  evidently  diameters  of  circles.  The  dimension 
9f  1"  is  the  diameter  of  the  circle  marked  a;  diameter  of  circle  b 
;",  of  circle  c,  9^1" ;  of  circle  d,  9£";  of  circle  e,  8",  and  of 
circle/,  6%".  Circles  b  and  c  are  dotted,  because,  imagining  the 
sectional  view  to  be  a  full  view,  when  the  ring  is  looked  at  from 
a  point  to  the  right  of  the  sectional  view,  all  that  part  of  it  to  the 
left  of  CC  is  hidden.  Further,  circles  b  and  c  are  imaginary, 
because  the  surface  they  are  supposed  to  indicate  is  a  curved 
surface;  still,  they  are  useful  in  helping  to  understand  the  draw- 
ing. The  fact  that  the  lines  CD  and  CD'  slope  shows  that  the 
ring  is  a  frustum  of  a  cone,  or  that  part  of  it  is  which  is  included 
between  DD'  and  CC.  That  part  included  between  circles  e  and 
g  is  flat,  and  extends  downward  to  form  a  cylindrical  ring,  whose 
inside  diameter  is  b\" ',  outside  diameter  is  8",  and  altitude  is 
tz  +  II  =  If".  Evidently,  there  is  a  groove  between  circles  e 
and  d;  there  is  also  another  groove  on  the  under  side,  as  in- 
dicated at  E  and  E' ,  which  are  at  the  bottom  of  another  conical 
surface.  According  to  the  note,  there  are  8  holes  spaced  equally, 
the  dotted  circle  showing  that  they  are  all  situated  at  the  same 
distance  from  the  center  o;  and  the  sectional  view  shows  that 
they  pass  clear  through  the  ring.  The  note  states  that  these 
holes  are  of  such  size  that  they  may  be  threaded  by  tapping 
with  a  yVth  inch  tap  having  12  threads  per  inch.  The  note  at 
the  bottom  of  the  drawing,  which  reads  "/"  all  over,  means 
that  the  ring  is  to  be  finished  all  over,  that  is,  no  part  of  it  is 
to  be  left  rough. 

13.  Standard  Sections. — The  sections  shown  in  Fig.  11  are 
practically  in  universal  use.  The  first  form,  shown  at  (a),  con- 
of  parallel  lines  spaced  at  equal  distances  apart,  all  lines 
being  of  the  same  weight:  this  form  is  always  used  for  cast  iron. 
The  sectioning  for  wrought  iron  is  shown  at  (6);  it  is  made  by 
drawing  light  and  heavy  lines  alternately,  parallel  and  equally  dis- 


§3 


REPRESENTING  SOLIDS  ON  PLANES 


19 


§§T<  is 


=±= 


3?  I   (S£ 


■ 


t-v3 


"^3 


u 


20 


HOW  TO  READ  DRAWINGS 


§3 


tant  apart.  The  sectioning  for  steel  is  shown  at  (c) ;  it  is  made  by 
drawing  pairs  of  parallel  lines,  the  distance  between  two  consecu- 
tive pairs  being  about  twice  the  distance  between  the  lines  form- 
ing the  pairs.  The  sectioning  for  brass  is  shown  at  (d);  it  is 
drawn  in  the  same  manner  as  the  sectioning  for  cast  iron,  except 
that  every  other  line  is  broken,  being  made  up  of  short  dashes. 
The  sectioning  for  wood  is  shown  at  (e) ;  the  upper  half  shows  the 


c 

ast  I 

ron 

(a) 


form  used  when  the  section  is  taken  across  the  grain,  and  the 
lower  half  is  used  when  the  section  is  taken  lengthwise,  or  with  the 
grain. 

There  are  other  forms  for  other  materials,  but  they  are  not 
much  used  and  there  are  no  other  recognized  standards,  different 
forms  being  used  to  indicate  the  same  material  in  different  draft- 
ing rooms.  The  ones  shown  in  Fig.  11  are  well  recognized  and 
are  in  nearly  universal  use. 

14.  When  the  surfaces  of  two  different  parts  come  together  in 
a  sectional  view,  the  section  lines  of  the  two  parts  are  made  to 
slope  in  different  directions  whenever  possible.  Thus,  referring 
to  Fig.  12,  which  represents  a  section  taken  through  a  bracket, 
which  holds  up  a  line  shaft,  and  its  bearing  P  is  the  bracket, 
S  is  the  shaft,  C  is  a  collar  that  is  forged  to  and  is  a  part  of 
the  shaft,  B  is  a  bushing  (which  can  be  removed  and  replaced  in 
case  of  vear),  and  R  is  a  collar  having  the  form  of  a  ring,  which  is 
held  in  place  by  a  set  screw  s  and  keeps  the  shaft  from  moving 
lengthwise.  As  shown  by  the  sectioning,  P  is  made  of  cast  iron, 
B  of  brass,  C  and  S  of  steel,  and  R  of  wrought  iron.  Note  that 
the  section  lines  for  P  and  B  and  for  R  and  P  slope  in  different 
directions;  it  was  not  possible  in  this  case  to  have  the  section 
lines  for  R  and  B  slope  in  different  directions,  but  since  they 
indicate  different  materials,  it  does  not  matter  particularly.  It 
will  also  be  noticed  that  the  lines  at  the  top  and  bottom  of  P 
are  ragged,  as  though  P  had  been  broken  off  at  these  places;  this 


^ 


iii-:rm-:sKNTiN<.;  solids  on  planes 


21 


is  exactly  what  the  draftsman  intended,  since  only  the  details 
of  the  bearing  are  to  be  brought  out.  For  the  same  reason, 
the  shaft  is  represented  as  though  broken  off  also;  whenever  a 


Fig.  12. 


round  piece  is  to  be  shown  as  broken  off,  it  is  always  drawn 
as  here  indicated.  Strictly  speaking,  the  shaft  should  also  have 
been  sectioned ;  but,  by  drawing  it  as  here  shown  and  sectioning 
the  ends  only,  the  drawing  is  more  readable  and  is  easier  to  make. 


I'l.;.    13. 

15.  Thin  Sections. — When  the  material  to  be  sectioned  is  very 
thin,  as  in  the  case  of  sheet  metal,  boiler  plates,  etc.,  or  when  the 
scale  used  is  so  small  that  the  edges  of  two  surfaces  in  a  sectional 
view  are  very  close  together,  it  is  then  the  custom  of  many  drafts- 
men to  make  the  surface  to  be  sectioned  entirely  black;  this  is  well 


HOW  TO  HEAD  DRAWINGS 


§3 


illustrated  in  Fig.  13.  which  shows  the  details  of  a  manhole  near 
one  end  of  a  boiler.  Here  A  is  the  boiler  shell,  shown  in  section  by 
the  solid  black  surface,  and  which  has  an  elliptic-ally  shaped  hole 
through  it.  as  indicated  in  the  other  view  and  by  the  lines  F;  B  is 
-•  1  ring  whose  inside  dimensions  are  the  same  as  those  of  the 
hole  in  the  shell,  the  ring  being  riveted  to  the  shell  in  order  to 


Section  on  AB 

Fig.  14. 

strengthen  it;  C  is  the  manhole  cover,  also  shown  in  solid  black 
sectioning,  which  is  drawn  up  against  the  inside  of  the  boiler 
and  held  in  place  by  the  two  studs  D'  and  D"  and  the  yokes  E' 
and  E",  as  clearly  shown  in  the  two  views.  Note  that  when  two 
black  sectioned  surfaces  come  together,  they  are  separated  by  a 
thin  white  line. 

16.  Conventional  Sections. — A  conventional  section  is  one 
that  is  not  drawn  in  strict  accordance  with  the  principles  of 
projection:  the  shaft  in  Fig.  12  and  the  rivets  in  Fig.  13  are  exam- 


§3  REPRESENTING  SOLIDS  ON  PLANES  23 

pies  of  conventional  sections.  Note  that  only  the  end  of  rivet 
6  is  shown;  this  shows  that  rivet  6  is  behind  the  cutting  plane, 
which  cuts  rivet  a. 

Another  example  of  a  conventional  section  is  shown  in  Fig. 
14.  At  first  glance,  it  would  appear  as  though  the  section  were 
taken  on  the  center  line  pq,  but  an  examination  of  the  top  view 
shows  that  if  it  were  not  for  the  part  CD  of  the  arms,  the  section 
would  appear  to  be  taken  on  the  center  line  mn.  As  a  matter  of 
fact,  the  sectional  view  has  been  drawn  as  though  the  part  that 
is  cross  hatched  were  a  section  on  mn,  and  the  arms  are  then 
drawn  in  between  the  cross-hatched  parts.  While  this  violates 
the  rules  of  projection,  the  resulting  drawing  is  clear,  and  this 
method  of  drawing  sectional  views  in  cases  similar  to  this  is 
almost  universal.  To  show  the  shape  of  a  cross  section  of  the  arm, 
it  is  drawn  on  the  arm  itself,  the  section  being  taken  on  the  line 
EF,  which  may  be  located  at  any  convenient  point  between  the 
hub  and  the  rim,  but  must  be  perpendicular  to  the  center  line  of 
the  arm .  Note  that  the  cross  section  of  the  arm  has  the  shape  of  an 
ellipse  and  that  the  arm  tapers,  the  larger  end  being  at  the  hub. 
As  the  drawing  has  been  made  to  a  very  small  scale,  the  shape  of 
a  section  of  the  rim  of  the  pulley  is  not  clearly  shown  in  the  sec- 
tional view;  hence,  a  section  is  taken  at  some  other  place,  as  AB, 
and  drawn  to  a  larger  scale.  This  section  must  be  a  radial 
section,  that  is,  the  trace  of  the  cutting  plane  must  coincide  with 
a  radius,  and  an  examination  of  the  section  shows  that  the  rim 
is  thicker  in  the  middle  than  at  the  ends,  the  difference  between 
the  two  being  called  the  crown  of  the  pulley.  Only  one  half  of 
the  pulley  is  shown  in  the  top  view,  the  lower  half  not  being 
necessary  because,  the  pulley  being  symmetrical,  the  lower 
half  is  exactly  the  same  as  the  upper  half. 


READING  DRAWINGS 


VISUALIZING  THE  OBJECT 


17.  A  drawing  serves  several  different  purposes:  it  serves  to 
preserve  in  permanent  form  the  shape  of  some  object,  and  to 
make  a  permanent  record  of  a  view  or  scene;  it  is  useful,  perhaps 
even  necessary,  in  explaining  or  in  understanding  the  details  of  a 
complicated  machine,  the  relations  of  the  various  parts  and  their 


24  HOW  TO  READ  DRAWINGS  §3 

operation.  If  properly  dimensioned,  the  drawing  may  be  used  to 
show  the  sizes  of  the  different  parts  for  purposes  of  comparison 
or  for  the  building  of  the  part  or  of  the  complete  machine.  In 
connection  with  the  first  purpose  mentioned  it  may  be  remarked 
that  drawings  have  been  found  in  France  on  the  walls  and  ceil- 
ings of  caves,  which  were  made  from  50,000  to  100,000  years  ago. 
These  drawings  are  very  crude,  but  are  nevertheless  intelligible 
and  useful,  representing  as  the}*  do  animals  that  were  extinct 
before  the  beginning  of  history. 

For  a  drawing  to  be  useful,  it  must  be  so  made  that  the  object 
or  scene  depicted  or  drawn  can  be  visualized;  that  is,  after  look- 
ing at  the  drawing,  the  mind  must  be  able  to  picture  or  imagine, 
the  object  represented  in  the  same  manner  as  though  the  object 
itself  were  being  viewed  by  the  eye.  In  the  case  of  a  perspective 
drawing,  there  is  no  difficulty  in  visualizing,  because  the  draw- 
ing represents  the  object  or  scene  as  it  appears  to  the  eye  when 
the  eye  is  in  some  chosen  position  relative  to  the  object,  the  position 
chosen  being  selected  by  the  draftsman.  Projection  drawings  are 
frequently  difficult  to  visualize,  but  they  can  always  be  pictured 
in  the  mind's  eye  if  the  reader  has  a  knowledge  of  the  principles 
heretofore  explained  and  exercises  sufficient  patience.  While 
progress  may  be  slow  at  first,  continued  practice  will  make  one 
proficient  in  reading  a  drawing.  It  takes  a  great  deal  of  practice 
to  associate  the  letters  of  a  word  so  that  the  word  is  recognized 
■>n  as  the  eye  sees  it;  it  is  the  same  way  with  a  drawing,  and 
speed  in  reading  a  drawing  comes  only  with  practice.  There  is 
no  reason  for  discouragement  if.  at  first,  the  reader  finds  it  diffi- 
cult to  visualize  an  object  represented  by  a  drawing;  reading 
drawings  is  an  art  that  can  easily  be  acquired  by  application  and 
with  practice. 

18.  When  reading  a  drawing,  several  facts  and  principles  should 
be  continually  kept  in  mind;  after  a  little  practice,  they  will  re- 
quire no  mental  effort  to  remember  them.  One  of  the  most 
important  of  these  principles  is  that  any  plane  (flat)  figure,  no 
matter  what  its  shape,  is  projected  on  any  cutting  plane  that 
intersects  the  surface  at  right  angles  as  a  right  line;  in  other 
words,  a  plane  figure  is  projected  into  the  trace  of  its  plane  as 
a  right  fine.  For  example,  in  Fig.  2,  the  plane  of  the  top  of  the 
frustum  cuts  the  picture  plane  in  the  line  c6.  which  is  the  trace 
of  the  plane  AC  on  the  picture  plane  P.  The  figure  A  BCD  is 
projected  into  the  trace  of  its  plane  on  plane  P  as  the  right 


§3  READING  DRAWINGS 


25 


line  cb,  and  the  circle  within  the  figure  ABCD  is  also  projected 
into  the  right  line  cb.  Again,  a  plane  figure,  no  matter  what  its 
shape,  is  projected  as  a  similar  and  equal  outline  on  any  plane 
parallel  to  the  plane  of  the  figure.  For  example,  in  Fig.  3,  the 
plane  N  is  parallel  to  the  plane  of  the  top  of  the  frustum,  and  the 
projection  of  the  top  on  plane  N  is  a'b'c'd',  which  is  equal  in  all 
respects  to  ABCD;  the  projection  of  the  circle  is  also  equal  to  the 
circle  (hole)  on  the  top  of  the  frustum. 

A  cylindrical  surface  is  projected  on  a  plane  parallel  to  one  of 
its  elements  as  a  quadrilateral,  and  the  projection  on  a  plane  per- 
pendicular to  an  element  has  the  shape  of  a  right  section  of  the 
cylindrical  surface.  If  the  cylinder  is  a  right  cylinder,  the  pro- 
jection on  a  plane  parallel  to  an  element  is  a  rectangle,  and  if  a 
right  section  is  a  circle  or  an  ellipse,  the  projection  on  a  plane 
perpendicular  to  an  element  is  a  circle  or  an  ellipse  also. 

Always  remember  that  every  point  on  the  object  or  within  the 
object  is  projected  into  some  point  on  every  complete  view  of  the  ob- 
ject, and  a  point  on  any  view  may  be  the  projection  of  one  or  of  any 
number  of  points  on  the  object.  Thus  in  Fig.  3,  the  point  c'"  is 
not  only  the  projection  of  the  point  C  but  also  of  D  and  of  any 
point  on  the  right  line  joining  C  and  D;  in  fact  c'"  is  the  projec- 
tion of  the  entire  line  CD,  because  CD  is  perpendicular  to  the 
plane  of  projection  KR.  If  the  line  joining  C  and  D  were  not 
straight,  but  were  a  plane  curve,  its  projection  on  plane  P  would 
be  a  right  line,  and  if  it  were  shaped  like  a  screw  thread,  its 
projection  on  plane  P  would  be  a  circle. 

It  is  therefore  evident  that  it  is  not,  in  general,  possible  to 
determine  from  one  view  what  any  point  or  line  represents;  but 
two  views— one  preferably  at  right  angles  to  the  other— will 
usually  be  sufficient,  and  three  views— one  being  perpendicular  to 
the  other  two— will  always  be  sufficient  to  determine  the  shape  of 
any  object.  If  there  is  any  doubt  as  to  what  any  particular  point 
on  a  view  represents,  draw  (or  imagine  as  drawn)  a  projector 
from  the  point  to  the  other  view;  if  the  projector  coincide  with  a 
line  in  the  other  view,  the  point  is  the  projection  of  that  line,  and 
the  point  is  also  the  projection  of  the  points  of  intersection  of  the 
projector  with  any  line  that  it  cuts  in  the  other  view. 

19.  Bearing  the  foregoing  in  mind,  refer  to  Fig.  4.  Looking 
first  at  the  plan  (top  view),  it  is  seen  that  it  consists  of  two 
rectangles  arranged  symmetrically,  one  within  the  other,  whose 
corners  are  connected  by  the  right  lines  a'h',  b'g',  etc.     There  is 


26  HOW  TO  READ  DRAWINGS  §3 

n< tilling,  however,  in  this  view  that  shows  whether  the  surfaces 
outlined  by  these  rectangles  are  flat  or  otherwise,  whether 
a'b'c'd'  is  parallel  to  h'g'f'e'  or  not,  or  whether  a'b'c'd'  is  above 
or  below  h'g'f'e';  there  is  also  nothing  to  show  whether  the  circle 
represents  a  hole  extending  from  a'b'c'd'  downward  or  whether 
it  represents  a  cylinder  extending  upward.  All  these  questions 
are  answered  by  referring  to  the  other  two  views.  Considering 
the  front  elevation,  plane  M ,  point  c"  is  the  projection  of  c'  or  b', 
point  d"  is  the  projection  of  d'  or  a',  and  line  d"c"  is  the  projection 
of  d'c'  or  a'b'.  Since  d"c"  is  the  only  line  between  d"  and  e" 
extending  across  the  figure,  it  is  the  projection  of  both  d'c'  and 
a'b',  c"  is  the  projection  of  the  entire  line  c'b',  and  d"  is  the  pro- 
jection of  the  entire  line  d'a'.  Since  both  these  lines  are  straight, 
it  follows  that  a'b'c'd'  is  a  flat  surface  perpendicular  to  the  plane  M. 
For  similar  reasons,  h'g'f'e'  is  also  a  flat  surface  perpendicular  to 
the  plane  M.  The  two  surfaces  are  therefore  parallel,  and  their 
distance  apart  is  the  perpendicular  distance  between  the  lines 
d"c"  and  e"f".  The  dotted  rectangle  shows  that  the  circle  in 
the  plan  represents  a  round  hole,  whose  depth  is  the  length  of 
i"j",  plane  M ,  or  i'"j'",  plane  P.  The  only  feature  still  undeter- 
mined is  the  character  of  the  lines  joining  the  corners,  that  is, 
the  lines  a'h',  b'g',  etc.  Since  these  lines  are  straight  in  all  three 
views,  they  are  straight  on  the  object  also,  and  the  sides  of  the 
object  are  likewise  flat.  Consequently,  the  object  is  either  a 
frustum  of  a  pyramid  or  a  rectangular  prismoid,  and  it  has  a 
hole  in  the  center  extending  from  the  smaller  base  to  within 
about  one-half  the  distance  between  the  bases.  The  actual 
position  of  the  hole  cannot  be  determined  by  observation  only, 
unless  the  views  are  fully  dimensioned.  Without  dimensions, 
the  position  of  the  hole  can  be  determined  only  by  the  use  of 
spacing  dividers  or  a  scale. 

20.  It  is  a  great  help,  sometimes,  to  imagine  the  paper  to  be 
folded  on  the  front  and  side  traces  JK  and  KL,  Fig.  4,  or  on  the 
front  and  vertical  side  traces  JK  and  KI,  Fig.  5,  so  that  planes 
M  and  P  will  both  be  perpendicular  to  plane  N,  as  in  the  glass 
box.  Then  a  projector  from  c"  will  intersect  a  projector  from 
c'"  passing  through  c',  thus  locating  C  on  the  object;  in  this 
way,  points  on  the  object  may  be  located  from  the  three 
views. 

Suppose  that  the  object  were  not  a  frustum  of  a  pyramid,  but 
that  the  top  view  were  the  same  as  before,  the  top  being  flat 


§3 


READING  DRAWINGS 


27 


and  sloping  so  that  all  the  edges  have  different  lengths.  A 
perspective  of  such  an  object  is  shown  in  Fig.  15  at  (a).  A  pro- 
jection drawing  giving  three  views  is  shown  in  the  same  figure. 
Note  that  the  projectors  have  been  omitted,  as  is  the  case  in  all 
practical  drawings.  To  read  this  drawing,  note  first  the  point  a", 
the  highest  point  in  the  elevation;  this  corresponds  to  a'  in 
the  plan,  because  if  it  corresponded  to  d' ,  a'  would  be  hidden  in 
the  front  view,  and  a"  d"  would  be  a  dotted  line  instead  of  a 


Fiu.   15. 


full  line.  For  the  same  reason,  b"  corresponds  to  &',  and  a"b" 
is  the  projection  corresponding  to  a'b';  similarly,  d" c"  is  the 
projection  corresponding  to  d'e',  d"a"  corresponds  to  d'a',  and 
c"b"  corresponds  to  c'b' .  The  quadrilateral  a"b"c"d"  is  the 
projection  of  the  top  of  the  object  on  the  front  plane.  By 
reasoning  in  a  similar  manner,  it  will  be  apparent  that  the 
quadrilateral  a"'V"d"&'"  is  the  projection  of  the  top  of  the 
object  on  the  right  side  plane.  Further,  since  the  bounding  lines 
of  these  projections  in  all  three  views  are  right  lines,  the  surface 


28  HOW  TO  READ  DRAWINGS  §3 

is  flat.  The  fact  that  the  surface  h'g'j'e'  is  projected  in  the 
right  lines  e"j"  and  j'"g'"  in  the  front  and  side  views  indicates 
that  this  surface  is  also  flat,  and  since  both  these  lines  are  horizon- 
tal, the  surface  represented  by  h'g'j'e'  is  a  plane  surface  perpen- 
dicular to  both  side  planes  and,  consequently,  parallel  to  the 
bottom  plane.  The  object,  therefore,  has  the  shape  of  a  prismoid 
that  has  been  cut  off  at  the  smaller  end  by  a  plane  making  an 
angle  with  all  three  planes  of  projection.  By  measurement  or 
with  the  spacing  dividers,  it  will  be  found  the  distance  of  b" 
from  e"j"  and  of  d"  from  e"j"  are  equal;  hence,  the  points  D  and 
B  on  the  object,  see  (a)  of  the  figure,  are  at  the  same  height 
above  the  plane  of  the  bottom  of  the  object. 


SOME  EXAMPLES  IN  READING  DRAWINGS 

21.  Guard  for  Movable  Jaw  of  Vise. — Fig.  16  shows  a  guard 
for  the  movable  jaw  of  a  vise.  The  jaw  is  made  of  cast  iron, 
while  the  guard  is  made  of  steel  and  forms  a  bearing  surface  for 
the  head  of  the  screw  that  moves  the  jaw.  Three  views  are 
given:  a  front  view,  a  bottom  view  (an  inverted  plan),  and  a 
sectional  side  view.  The  center  run  divides  the  guard  into  two 
equal  parts  and  is  an  axis  of  symmetry.  For  convenience,  the 
same  letters  of  reference  have  been  used  on  both  halves  about 
the  center  line  mn.  In  what  follows  the  front  view  will  be 
designated  as  (A),  the  bottom  view  as  (B),  and  the  sectional  side 
view  as  (C). 

The  center  line  mn  is  supposed  to  be  perpendicular  to  the  bottom 
plane,  and  is  therefore  parallel  to  the  front  and  side  picture 
planes.  The  line/7'  in  (B)  is  the  projection  of  fa,  the  semicircle 
aaa,  and  aj;  it  is  also  the  projection  of  jc,  rr,  and  cj;  hence,  the 
front  of  the  guard,  as  represented  by  fcrrcfaa  a  is  flat  and  per- 
pendicular to  the  bottom  picture  plane,  as  also  indicated  by  the 
line  r'"  f"  in  (C).  The  semicircle  bbb  is  projected  in  the  right 
line  b'b'  in  (B)  and  is  connected  to  the  semicircle  aaa  by  a 
circular  arc  having  a  radius  of  If  in.  This  arc  is  also  shown 
in  (C),  and  has  the  same  radius.  Consequently,  the  surface 
represented  by  ab,  avbv,  and  a'b'  is  a  curved  surface,  and  since  its 
portions  on  three  planes  perpendicular  to  one  another  are  circular 
arcs,  it  is  a  surface  of  revolution;  it  is  not  a  part  of  a  spherical 
surface,  because  the  radius  of  a  section  perpendicular  to  the  axis 


§3 


READING  DRAWINGS 


29 


is  not  the  same  as  for  a  section  that  includes  the  axis,  the  radius 
of  the  former  being  1  j  in.  and  of  the  latter  If  in.  To  make  this 
clear,  the  head  of  the  screw,  which  bears  against  this  surface,  is 
shown  in  detail  at  (a).  The  thickness  of  the  guard  at  the  point  b 
is  indicated  by  the  length  of  s'b'  and  svbv  in  {B)  and  (C).  The 
edge  cr  is  straight  and  vertical  and  is  intersected  by  the  cylindrical 
surface  of  which  fc  is  a  part;  the  intersection  is  denoted  by  the 
point  c  in  (A),  by  the  dotted  line  c'd'  in  (B),  and  by  the  dotted 
line  c"d"  in  (C).     The  points  a  and  c  are  both  projected  into  the 


Ui) 


,/R       '/  4' in 
lltt        ;     j'dttr1 


■-:V/    !  ~T~^y 


same  point  on/'/'  in  (5)  and  into/"  in  (C),  because  they  both  lie 
on  the  same  surface  and  are  at  the  same  distance  from  the  center 
line  mn.  On  the  back  of  the  guard,  there  is  a  shoulder  on  either 
side  of  the  center  line,  as  indicated  by  the  dotted  lines  ihge)  as 
neither  shoulder  is  touched  by  the  cutting  plane,  the  shoulder 
is  not  sectioned  in  (C),  but  its  outline  is  indicated  by  i"h"g"e"j"l", 
the  line  g"j"  in  (C)  and  g'f  in  {B)  being  the  projection  of  the  line 
of  intersection  of  the  flat  surface  kg  and  the  cylindrical  surface 
ge.  The  dotted  line  in  the  hole  in  ((')  is  the  edge  that  extends 
backward  from  r.  The  line  e'k'  in  (B)  is  the  bottom  edge 
(inside)   of  the  shoulder;  it  is  projected  into  the  point  e"  in 


30  HOW  TO  READ  DRAWINGS  §3 

(C);  this  fad  shows  thai  the  bottom  en. Is  of  the  guard  are  flat, 
the  outline  of  the  surface  being  fa'b's'e'hfV. 

To  assist  in  visualizing  the  guard,  a  perspective  of  it  is  shown 
in  (6).  By  comparing  the  lines  on  the  three  views  with  the  cor- 
responding lines  on  the  perspective,  the  reader  should  have  no 
difficulty  in  forming  a  mental  picture  of  this  object.  As  a 
perspective  is  seldom  given,  the  reader  should  make  every  effort 
to  visualize  from  the  three  views,  without  the  aid  of  the  perspec- 
tive. The  three  views  will  show  him  the  shape  of  the  front,  the 
shape  of  the  back,  the  shape  of  the  top  and  bottom,  and  the  shape 
of  the  bearing  surface;  knowing  these,  it  will  then  be  easy 
to  visualize  the  entire  object. 

22.  The  Mercer  Cell.— Fig.  17  shows  a  Mercer  cell,  which  is 
used  for  the  production  of  caustic  soda  and  chlorine  by  elec- 
trolysis. A  top  view  of  the  cell  is  shown  at  (a),  and  a  cross 
section  on  the  line  AB  is  shown  at  (b).  As  indicated  by  the 
sectioning,  the  cell  is  an  earthenware  crock;  it  is  open  at  the 
top  and  has  five  rectangular  openings  a  in  the  cylindrical  part. 
There  is  also  a  cover,  shown  at  (c),  which  fits  into  the  groove  b. 
Attention  is  called  to  the  funnel-shaped  part  d,  which  is  a  part  of 
the  crock;  its  shape  will  be  readily  perceived  with  the  aid  of  the 
top  view  (a).  The  cylindrical  part  of  the  crock  is  covered  with  a 
layer  of  asbestos,  extending  to  just  above  the  top  of  the  rectan- 
gular openings;  this  is  held  in  place  by  perforated  sheet  iron, 
which  acts  as  an  electrode,  and  which  is  kept  in  position  by  three 
iron  bands  /,  the  latter  being  made  in  three  pieces,  flanged,  and 
bolted,  as  shown  at  (c)  and  (d).  Inside  the  crock  are  placed  four 
carbon  electrodes,  which  are  attached  to  the  disk  h,  as  shown  at 
(e);  this  view  shows  only  two  electrodes,  the  other  two  being 
behind  the  two  that  are  seen.  The  disk  h  rests  on  the  lugs  c, 
shows  in  views  (a)  and  (o).  At  (/)  is  shown  the  cover,  which  fits 
into  the  groove  b.  The  projection  j  serves  not  only  as  a  handle 
but  also  as  an  outlet  through  which  the  lead  i,  attached  to  the  disk 
k  at  its  center,  passes.  The  lead  i  is  connected  to  the  generator, 
and  it  conducts  the  current  from  the  generator  to  the  carbons  g. 
The  chlorine  gas  escapes  through  the  outlet  k,  to  which  is 
attached  the  pipe  I  that  conducts  the  gas  to  the  main,  see  (c). 

An  elevation  of  the  entire  arrangement  is  shown  at  (c),  and  a 
battery  of  six  of  these  cells  placed  in  a  sheet-iron  tank  is  shown  at 
(g).  This  drawing  is  very  simple,  and  requires  no  further 
explanation. 


§3 


READING  DRAWINGS 


31 


FlQ.     17. 


32 


HOW  TO  READ  DRAWINGS 


§3 


§3  READING  DRAWINGS  33 

23.  Rotary  Drying  Furnace. — Fig.  18  shows  in  a  general  way 
the  manner  in  which  the  black  liquor  in  the  soda  process  is  evapor- 
ated and  the  black  ash  is  obtained  in  the  recovery  of  the  soda. 
The  furnace  a  is  mounted  on  wheels  so  it  can  be  moved  away  from 
the  rotary  incinerator  b,  for  cleaning  and  repairs.  The  incinerator 
is  a  horizontal  steel  cylinder,  lined  with  fire  brick,  as  indicated  in 
the  sectional  view  (a),  which  is  a  section  on  the  line  AB  looking 
toward  the  right,  as  indicated  by  the  arrows.  Steel  tires  c  are 
attached  to  the  incinerator,  and  these  rest  on  bearing  wheels  d, 
the  wheels  supporting  the  entire  weight  of  the  incinerator.  The 
incinerator  is  caused  to  revolve  by  means  of  the  gear  and  pinion 
arrangement  e,  the  pinion  being  driven  by  a  system  of  compound 
gears,  which,  in  turn  are  driven  by  the  engine  m,  as  indicated  in 
the  sectional  view.  The  thick  liquor  is  stored  in  the  tank/,  from 
which  it  flows  through  the  pipe  g  and  nozzle  h  into  the  rear  end  of 
the  incinerator  through  the  opening  j.  The  front  end  is  also 
open;  and  as  the  ash  is  formed,  it  gradually  accumulates,  and 
falls  out  through  this  opening  into  a  pit  or  into  a  car  below  it. 
The  hot  gases  pass  through  the  incinerator,  through  the  opening 
j,  under  and  through  the  boiler  k  (heating  the  water  therein),  and 
through  the  stack  I  into  the  atmosphere.  The  details  of  the 
process  should  now  be  clear  without  further  explanation. 

24.  Worm  Washer. — Fig.  19  shows  a  form  of  apparatus  used 
for  washing  wood  pulp.  Two  views  are  given,  (6)  being  an  end 
view  and  (a)  a  longitudinal  section  taken  on  the  line  AB.  The 
apparatus  consists  of  two  perforated  metal  cylinders  a,  which  are 
partly  enclosed  in  a  wooden  box  or,  tank,  j.  The  cylinders 
have  at  one  end  a  tire  d,  which  runs  between  the  flanges  of  the 
bouble-flanged  wheels  e.  At  the  other  end,  the  cylinder  is  sup- 
ported by  the  hollow  trunnion  c,  which  turns  in  the  bearing  p. 
The  cylinder  is  thus  supported  by  the  wheels  e  at  one  end  and  the 
hollow  trunnion  at  the  other  end.  By  referring  to  view  (6), 
it  will  be  noted  that  the  box  is  divided  into  two  compartments  at 
the  end  k,  which  is  funnel-shaped.  Note  particularly  the  shape  of 
this  partition,  indicated  by  the  dotted  Ymcabcdefgh  in  (6), apart 
of  it  being  circular  to  permit  the  cylinder  to  turn  in  it,  and,  at  the 
same  time,  follow  the  outline  of  the  cylinder.  The  cylinder  being 
thin,  heavy  iron  bands  o  encircle  it  at  intervals,  in  order  to 
strengthen  it.  In  both  cylinders  is  a  sheet-metal  worm  b,  which 
is  attached  to  the  head  of  the  closed  end  of  the  cylinder  that 
carries  the  hollow  trunnion.     At  the  end  of  the  trunnion,  is  a 


34 


HOW  TO  READ  DRAWINGS 


§3 


§3  READING  DRAWINGS  35 

large  sprocket  wheel/;  these  sprocket  wheels  are  connected  by  an 
endless  belt  q,  so  that  both  cylinders  may  revolve  together.  A 
perforated  pipe  h  extends  along  the  entire  length  of  both  cylin- 
ders, and  is  closed  at  the  farther  end.  The  operation  of  the 
apparatus  is  as  follows:  The  stock  to  be  washed  flows  through  the 
pipe  g  and  hollow  trunnion  c  into  the  cylinder,  which,  as  it 
slowly  turns,  carries  it  through  the  cylinder  by  means  of  the 
worm.  Water  flows  through  the  holes  in  the  perforated  pipe  h 
and  enters  the  cylinder  through  its  perforations,  washing  the 
stock  and  falling  through  the  holes  into  the  open  space  in  the 
box  below  the  cylinder,  from  whence  it  is  discharged  through  the 
pipes  m.  The  washed  stock  from  the  cylinders  falls  into  the 
compartments  k,  and  is  washed  into  the  exit  pipe  I  by  water 
discharged  from  the  pipes  i.  The  slanting  boards  n  keep  the 
water  from  splashing  out  of  the  box. 

25.  Remark. — The  object  of  the  last  two  figures  is  to  illustrate 
the  application  of  the  apparatus  described;  consequently,  the 
drawings  fail  to  show  many  of  the  details  that  would  be  necessary 
to  reproduce  the  apparatus.  Drawings  of  this  kind  are  a  feature 
of  printed  matter — books,  technical  journals,  catalogs,  etc. — 
when  the  main  object  of  the  descriptive  matter  is  to  explain 
processes,  the  operation  of  machines,  etc. 

Assuming  that  the  reader  has  carefully  studied  all  that  precedes 
this  article  and  has  conscientiously  compared  all  letters  of  refer- 
ence with  the  cuts,  he  ought  now  to  be  able  to  read  intelligently 
most  of  the  drawings  that  he  is  likely  to  encounter.  Complicated 
ones  may  give  him  trouble,  not  from  lack  of  knowledge,  but  from 
lack  of  practice  in  reading  drawings.  He  will  find,  at  times, 
that  a  knowledge  of  the  application  or  purpose  of  the  object 
drawn  is  necessary  to  a  complete  understanding  of  the  drawing,  as 
in  Fig.  19,  for  instance.  In  any  case,  he  ought  to  be  able  to  read 
any  drawing  or  to  understand  any  illustration  given  in  this 
course. 


SECTION  4 

ELEMENTS  OF  PHYSICS 

(PART  1) 


MATTER,  MOTION,  AND  FORCE 


MATTER  AND  ITS  PROPERTIES 

1.  Definition. — The  word  matter,  as  used  in  its  scientific 
sense,  means  anything  that  occupies  space.  Since  any  finite 
portion  of  space  has  length,  breadth  (or  width),  and  thickness 
(or  height  or  depth),  it  can  be  measured;  and  matter,  which 
occupies  a  certain  space,  can  also  be  measured,  and  in  the  same 
units.  As  will  shortly  be  shown,  matter  can  also  be  measured 
in  units  other  than  those  used  to  measure  space. 

This  property  of  matter,  that  it  occupies  space,  is  called 
extension,  and  it  is  a  property  (or  characteristic)  common  to  all 
matter.  Any  property  that  is  common  to  everything  is  called  a 
general  property  of  that  thing;  hence,  extension  is  a  general 
property  of  matter. 

2.  General  Properties  of  Matter.- — When  it  is  desired  to  speak 
of  a  certain  amount  of  matter  without  specifying  what  it  is,  iron, 
water,  air,  clay,  etc.,  it  is  called  a  body.  A  body  may  be  of  any 
size — so  small  as  not  to  be  visible  or  as  large  as  the  earth  or  larger 
— it  simply  means  a  certain  amount   of  matter. 

In  addition  to  extension,  matter  possesses  a  number  of  other 
general  properties:  impenetrability,  divisibility,  indestructi- 
bility, inertia,  porosity,  compressibility,  expansibility,  elasticity, 
mobility,  and  weight.  These  properties  are  common  to  all 
matter  and,  consequently,  to  all  bodies.  A  thorough  under- 
standing of  the  meaning  of  these  words  will  be  a  great  help  in 
understanding  physics  and  chemistry. 

3.  Three  Forms  of  Matter. — All  matter  may  exist  in  one  of 
three  forms  or  states:  the  solid,  the  liquid,  or  the  gaseous  state. 

1 


2  ELEMENTS  OF  PHYSICS  §4 

A  solid  body  is  one  that  retains  its  shape  under  ordinary  condi- 
tions; thus,  wood,  stone,  clay,  iron,  etc.  are  solids  under  ordinary 
conditions. 

A  liquid  body  (or  liquid)  is  one  that  does  not  keep  its  shape 
unless  placed  in  a  vessel  to  keep  it  from  spreading.  It  is  charac- 
teristic of  all  liquids  that  they  spread  out  and  conform  to  the  shape 
of  the  vessel  that  contains  them.  Further,  if  the  upper  surface 
is  free,  not  touched  by  any  solid,  it  will,  under  ordinary  condi- 
tions, be  a  flat,  level  surface — for  all  practical  purposes,  a  plane 
perpendicular  to  a  diameter  of  the  earth  passing  through  the 
center  of  the  surface. 

A  gaseous  body  (or  gas  )  is  one  that  completely  fills  any  closed 
vessel  that  contains  it.  If  the  vessel  be  connected  to  another 
closed  vessel,  say  by  a  pipe  having  a  cock,  so  that  communication 
may  be  opened  or  closed  between  the  two  vessels,  and  the  cock 
be  opened,  the  gas  will  fill  both  vessels  completely;  and  any 
particular  amount  of  space  in  either  vessel,  say  1  cubic  inch,  will 
contain  exactly  the  same  weight  of  gas;  but  the  weight  of  a  cubic 
inch  of  the  gas  will  not  be  the  same  as  in  the  first  vessel  before 
the  cock  was  opened. 

Most,  if  not  all,  substances  can  exist  in  all  three  states;  thus, 
water  under  ordinary  conditions  is  a  liquid,  when  frozen  it  is  a 
solid  (ice),  and  when  vaporized  it  is  a  gas  (steam).  The  same  is 
true  of  iron,  which  can  be  melted  (liquefied)  and  vaporized,  and 
of  most  other  substances.  Obviously,  a  substance  can  exist  in 
but  one  state  at  a  time. 

4.  Divisibility. — Divisibility  means  that  an3r  bod}^  may  be 
divided  into  two  or  more  bodies.  A  solid  may  be  divided  by  cut- 
ting or  by  macerating  or  pulverizing  it;  a  liquid  may  be  divided 
by  pouring  some  of  it  into  another  vessel;  a  gas  may  be  divided 
as  described  in  the  last  article.  Liquids  and  gases  are  frequently 
referred  to  under  the  general  name  of  fluid  bodies  or  fluids; 
hence,  a  fluid  may  be  either  a  liquid  or  a  gas.  There  are,  of 
course,  other  ways  of  dividing  bodies  than  those  here  mentioned. 

5.  Molecules  and  Atoms. — If  an  ounce  of  salt  be  put  into  a 
pound  of  water,  the  salt  will  apparently  disappear;  the  salt, 
however,  is  still  in  the  water,  as  can  be  proved  by  various  tests, 
as,  for  instance,  by  tasting.  Here  the  water  is  said  to  dissolve 
the  salt.  What  really  occurred  was  that  the  water  divided  the  salt 
into  particles  so  small  that  they  were  no  longer  visible.     By 


§4  MATTER,  MOTION,  AND  FORCE  3 

stirring  the  water,  the  salt  particles  may  be  made  to  occupy 
every  part  of  the  water.  If  more  water  is  added,  the  salty  taste 
becomes  less  strong,  and  finally  becomes  indistinguishable;  but 
the  salt  is  still  there,  though  it  has  been  divided  into  extremely 
small  particles.  In  accordance  with  modern  theory,  the  division 
cannot  be  carried  on  indefinitely,  for  when  a  certain  stage  has  been 
reached,  any  further  division  will  change  the  nature  of  the  body. 
For  example,  it  is  shown  in  chemistry  that  salt  is  a  compound 
of  sodium  and  chlorine,  one  part  of  each.  When  the  division 
reaches  a  certain  point,  the  particles  can  no  longer  be  divided  by 
any  of  the  means  heretofore  mentfoned,  and  they  are  then  called 
molecules.  If  a  molecule  be  divided  by  chemical  action  or  by 
the  action  of  electricity  or  heat,  the  parts  are  called  atoms.  In 
the  case  of  salt,  the  molecule  when  divided  becomes  two  atoms, 
one  being  sodium  and  the  other  chlorine.  It  has  not  been  found 
possible  so  to  divide  an  atom,  and  it  is  therefore  assumed  that  an 
atom  is  indivisible.  Atoms  do  not  usually  exist  in  a  free  state, 
two  or  more  being  united  to  form  a  molecule. 

A  molecule,  then,  may  be  defined  as  the  smallest  particle  of 
matter  that  can  exist  without  changing  the  nature  of  the  sub- 
stance of  which  it  forms  a  part;  thus,  a  molecule  of  water  consists 
of  two  atoms  of  hydrogen  and  one  atom  of  oxygen,  and  if  a 
molecule  of  water  be  divided,  it  will  be  no  longer  water,  but  three 
atoms,  which  will  immediately  reunite  to  form  water  again  or 
unite  among  themselves  to  form  molecules  of  hydrogen  and 
oxygen.  Molecules  are  exceedingly  small,  so  small  that  they  have 
never  been  seen  and  probably  never  will  be  seen.  They  are  as- 
sumed to  be  spherical  in  shape  and  are  known  to  be  in  very  rapid 
motion,  moving  in  all  directions  and  constantly  colliding.  The 
molecules  of  a  solid  are  more  numerous  and  move  through  shorter 
distances  than  those  of  a  liquid,  and  those  of  a  liquid  are  more 
numerous  and  move  through  shorter  distances  than  those  of  a 
gas. 

6.  Porosity. — All  matter  is  porous,  that  is,  the  molecules  do  not 
completely  fill  the  space  occupied  by  the  molecules;  otherwise, 
there  could  be  no  movement  of  the  molecules.  Even  though 
they  may  not  be  visible  to  the  naked  eye  or  under  the  highest 
powered  microscope,  all  bodies  have  minute  pores  or  channels, 
as  is  proved  by  the  fact  that  water,  for  example,  can  be  forced 
through  a  highly  polished,  thin  steel  plate.  If  matter  were  not 
porous,  it  would  not  be  possible  to  force  light  or  fluids  through  a 


4  ELEMENTS  OF  PHYSICS  §4 

sheet  of  metal,  no  matter  how  thin  it  might  be;  but  a  piece  of 
gold  leaf  1 /300,000th  of  an  inch  thick  will  allow  light  to  pass 
through  it. 

7.  Impenetrability.- — Impenetrability  refers  to  that  property 
whereby  two  bodies  cannot  occupy  exactly  the  same  space  at  the 
same  time.  If  a  stone  be  placed  in  a  vessel  partly  filled  with 
water,  the  level  of  the  water  will  be  raised,  showing  that  the  stone 
has  taken  the  place  of  the  same  amount  of  water.  Advantage 
may  be  taken  of  this  fact  to  find  the  volume  of  any  irregular 
solid  that  is  not  dissovled  in  water.  For  instance,  knowing  the 
volume  of  the  water  before  and  after  the  solid  is  immersed  in  it, 
the  difference  will  be  the  volume  of  the  solid.  If  a  solid  be  dis- 
solved in  a  liquid,  the  volume  may  be  the  same  as  before,  as,  for 
example,  dissolving  a  certain  amount  of  sugar  in  a  cup  full  of  tea. 
This  does  not  mean,  however,  that  water  is  penetrable;  the  experi- 
ment simply  shows  that  water  is  porous,  and  that  the  molecules  of 
sugar  have  crowded  between  the  molecules  of  water  and  occupy 
some  of  the  space  not  filled  by  the  water  molecules.  Again,  if 
50  c.c.  (cubic  centimeters)  of  alcohol  are  mixed  with  50  c.c.  of 
water,  the  volume  of  the  mixture  will  not  be  100  c.c,  but  97 
c.c;  this  shows  that  one  of  the  liquids  is  more  porous  than  the 
other,  with  the  result  that  the  combined  volume  is  less  than 
the  sum  of  the  volumes  before  mixing. 

8.  Compressibility. — Compressibility  refers  to  that  property 
whereby  bodies  can  be  made  to  occupy  a  smaller  space.  Gases 
are  very  compressible;  liquids  and  solids  are  only  slightly  com- 
pressible, but  there  is  no  substance  that  is  incompressible.  This 
necessarily  follows  from  the  fact  that  matter  is  porous,  and  the 
further  fact  that  the  molecules  can  be  brought  closer  together 
and  be  made  to  occupy  a  smaller  space  than  before. 

9.  Expansibility. — Expansibility  refers  to  that  property  whereby 
the  molecules  can  be  forced  farther  apart  and  be  made  to  occupy 
a  greater  space.  All  gases  tend  to  expand  and  occupy  a  greater 
volume;  but  liquids  and  solids  can  usually  be  made  to  expand 
only  by  means  of  heat,  most  of  them  expanding  when  the 
temperature  is  raised,  although  a  few  expand  under  certain 
circumstances  when  the  temperature  is  lowered. 

10.  Elasticity.- — Elasticity  is  the  name  of  the  property  by 
which,  after  a  body  has  been  distorted  (shape  changed)  in  any 
way  and  by  any  means  (except  division)  the  body  tends  to  resume 


§4  MATTER,  MOTION,  AND  FORCE  5 

its  former  shape  when  the  cause  that  produced  the  distortion  is 
removed  and  the  former  conditions  are  restored.  While  the 
body  may  not  entirely  regain  its  former  shape,  it  always  has  a 
tendency  that  way  and  will  partly  regain  it.  A  piece  of  rubber, 
for  instance,  may  be  stretched,  and  if  not  stretched  too  much' 
will  resume  its  former  shape  when  released.  A  ball  of  moist 
clay  or  putty  may  be  pressed  and  made  to  assume  and  retain  any 
shape  but  it  does  not  retain  exactly  the  shape  it  had  when  the 
pressure  was  removed.  All  fluids  and  some  solids,  such  as  steel, 
ivory,  glass,  etc.,  are  very  elastic. 

11.  Mobility. — The  distance  between  one  body  and  another 
body  may  be  changed  by  moving  either  or  both  bodies.  This 
property  is  referred  to  as  mobility.  Thus,  no  matter  how  firm 
the  foundation  or  how  heavy  and  rigid  the  structure  on  it,  an 
earthquake  will  move  it.  The  earth  itself,  the  sun,  and  all  the 
stars  are  moving;  there  is  no  such  thing  as  an  immovable  body. 

12.  Inertia. — Inertia  means  without  life,  without  power  of 
self  movement.  As  applied  to  bodies  in  a  scientific  sense,  it 
means  that  if  a  body  is  at  rest,  it  cannot  put  itself  in  motion;  or  if 
it  is  in  motion,  it  cannot  bring  itself  to  rest  or  change  its  direction 
of  motion.     This  will  be  explained  more  fully  later. 

13.  Weight.— Every  body  exerts  a  certain  attraction  on  every 
other  body,  which  tends  to  make  the  bodies  come  together  and 
touch  one  another.  As  between  two  bodies  of  ordinary  size,  this 
attraction  is  very  small;  but  the  earth  is  so  large  that  this  attrac- 
tion between  the  earth  and  other  bodies  is  very  manifest,  and  the 
result  is  called  weight.  Every  solid  or  liquid  body,  no  matter 
how  small,  will  fall  to  the  earth  unless  sustained  by  some  inter- 
vening body.  Gases  are  also  attracted  by  the  earth,  even  though 
they  may  rise  and  go  away  from  it  when  released.  This  may  be 
proved  by  weighing  an  empty  vessel  in  a  vacuum  (a  place  where 
there  is  no  air  or  gas  of  any  kind),  then  filling  the  vessel  with  a  gas 
or  mixture  of  gases,  say  air,  and  weighing  again;  the  weight  in  the 
second  case  will  be  greater  than  in  the  first  case,  and  the  addi- 
tional weight  will  be  the  weight  of  the  air. 

14.  Indestructibility.— It  i  s  impossible  to  destroy  matter.  A 
body  may  be  separated  into  molecules  and  the  molecules  into 
atoms,  but  the  atoms  will  unite  to  form  other  molecules  and  other 
bodies,  and  there  will  be  the  same  number  of  atoms  as  before. 
Matter  may  assume  countless  forms  and  undergo  innumerable 


6  ELEMENTS  OF  PHYSICS  §4 

changes,  but  the  same  number  of  atoms  is  present  in  the  uni- 
verse ;  this  is  otherwise  expressed  as  the  principle  of  conservation 
of  matter,  and  it  means  that  matter  is  indestructible.  Matter 
may  be  transformed,  but  it  cannot  be  destroyed,  neither  can  the 
total  amount  of  matter  in  the  universe  be  changed. 

15.  Specific  Properties  of  Matter. — In  addition  to  the  general 
properties,  bodies  possess  certain  other  properties,  not  common  to 
all  bodies,  which  are  therefore  called  specific  properties.  Some 
of  the  most  important  of  these  specific  properties  are  rigidity, 
pliability,  flexibility,  malleability,  ductility,  tenacity,  brittle- 
ness,  and  hardness.  These  terms  may  be  defined  briefly  as 
follows : 

Rigidity  is  the  resistance  offered  by  a  body  to  a  change  in  its 
shape;  steel  is  very  rigid.  Pliability  is  the  ease  with  which  the 
shape  of  a  body  may  be  changed;  lead,  copper,  etc.  are  pliable. 
A  body  is  flexible  when  it  can  be  bent  without  breaking,  as  a 
spring,  a  rope.  Malleability  is  that  property  that  indicates  that 
the  body  possessing  it  may  be  rolled  or  hammered  into  sheets ; 
gold  is  the  most  malleable  of  all  known  substances.  Ductility 
is  that  property  of  the  body  possessing  it  which  indicates  that 
the  body  may  be  drawn  out  into  wire;  platinum  possesses  greater 
ductility  than  any  other  substance.  Tenacity  refers  to  the 
resistance  offered  by  some  bodies  to  being  pulled  apart;  steel  is 
extremely  tenacious.  Brittleness  is  the  term  used  to  indicate 
that  some  bodies  are  easily  broken  when  subjected  to  sudden 
shocks;  glass,  ice,  etc.  are  very  brittle.  Hardness  is  that  prop- 
erty that  indicates  that  a  body  possesses  the  ability  to  scratch 
some  other  body.  The  hardness  of  any  particular  body  may  be 
determined  by  using  it  to  scratch  some  other  body;  it  will  be 
harder  than  the  body  it  scratches  and  softer  than  any  body  that 
scratches  it.  The  diamond  is  the  hardest  substance  known;  no 
substance  will  scratch  a  diamond  except  another  diamond,  in 
which  case,  either  will  scratch  the  other. 


MOTION  AND  VELOCITY 

16.  Definition. — If  during  a  certain  space  of  time,  the  distance 
between  two  bodies  is  increasing  or  decreasing,  one  body  is  said  to 
be  in  motion  relative  to  the  other  body.  If  a  body  A  occupy  and 
continue  to  occupy  a  certain  position  on  the  earth's  surface,  A  is 
said  to  be  at  rest  (or  is  fixed)  relative  to  the  earth;  if,  now,  the 


§4  MATTER,  MOTION,  AND  FORCE  7 

distance  between  A  and  another  body  B  continually  varies 
during  a  time  t,  B  is  said  to  be  in  motion  relative  to  A,  or  B  has 
motion  relative  to  A. 

The  use  of  the  word  relative  in  the  last  paragraph  requires  a 
special  explanation.  All  motion  is  relative;  there  is  nothing  in 
space  that  is  absolutely  at  rest.  The  circumference  of  the  earth 
at  the  equator  is  about  25,000  miles,  and  it  turns  around  on 
its  axis  once  in  24  hours  or  86,400  seconds;  hence,  an  object 
on  the  equator  is  moving  around  with  the  earth  at  the  rate  of 

oFTnn =  1524  feet  every  second.     The  earth  is  also  moving 

in  its  orbit  about  the  sun  with  a  speed  of  about  18  miles  every 
second;  the  sun  is  also  in  motion,  etc.  Consequently,  when  it  is 
stated  that  a  body  is  at  rest,  the  usual  meaning  is  that  the  body 
is  at  rest  relative  to  a  point  on  the  earth's  surface;  and  if  a 
body  is  in  motion  relative  to  this  point,  it  is  moving  toward  or 
from  that  point. 

A  body  may  be  in  motion  relative  to  one  point  and  at  rest  rela- 
tive to  another,  both  at  the  same  time.  For  instance,  suppose  a 
person  to  be  on  a  slowly  moving  railway  train ;  if  he  walk  from  one 
end  of  the  train  to  the  other  at  the  same  rate  of  speed  that  the 
train  is  moving,  but  in  the  opposite  direction,  he  will  be  in  motion 
relative  to  a  fixed  point  on  the  train,  but  at  rest  relative  to  a  fixed 
point  beneath  him  on  the  earth.  In  other  words,  the  center  of 
his  body  will  remain  over  the  point  on  the  earth  until  he  reaches 
the  end  of  the  train. 

17.  Path  of  a  Body. — In  order  to  compare  motions,  it  is  as- 
sumed that  the  motion  is  the  same  as  that  of  a  very  small  particle 
of  the  body  at  the  center  of  gravity  of  the  body;  and  the  line 
described  by  this  particle  is  called  the  path  of  the  body.  The 
path  may  be  straight  or  curved,  but  the  length  of  the  path  is 
always  equivalent  to  a  right  line  having  the  same  length.  If  the 
path  is  a  right  line,  the  direction  of  motion  is  along  that  line,  i.e., 
along  the  path;  but  if  the  path  is  a  curve,  the  direction  of  motion 
is  along  a  tangent  to  the  path  at  the  point  occupied  by  the  body 
(center  of  the  body)  at  the  instant  considered.  This  will  be 
explained  more  fully  in  the  section  on  mechanics. 

18.  Velocity. — Velocity  is  rate  of  motion.  If  the  speed  of  a 
body  is  uniform,  that  is,  if  the  body  passes  over  equal  distances  in 
equal  times,  the  velocity  is  equal  to  the  distance  divided  by  the 


8  ELEMENTS  OF  PHYSICS  §4 

time.     Thus,  if  v  =  the  velocity,  8  =  the  distance,  and  t  =  the 

time, 

s 

V  =  t 

The  unit  of  velocity  is  a  compound  unit,  because  it  requires 

more  than  one  unit  to  express  it — a  unit  of  distance  (length)  and  a 

unit  of  time.     If  s  is  measured  in  feet  and  t  in  minutes,  the  unit  of 

velocity  is  one  foot  divided  by  one  minute,  and  is  called  one  foot 

per  minute,  the  word  per  meaning  divided  by;  whenever  the  word 

per  occurs  in  the  name  of  a  compound  unit,  it  always  has  this 

meaning.     If  s  is  measured  in  miles  and  t  in  hours,  the  unit  of  v  is 

one  mile  per  hour;  and  if  s  is  in  centimeters  and  t  in  seconds,  v  is  in 

centimeters  per  second. 

19.  When  a  body  passes  through  equal  distances  in  equal 
times,  its  velocity  is  said  to  be  uniform  or  constant  otherwise  it  is 
variable.  But  whether  uniform  or  variable,  the  velocity  of  a 
body  at  any  instant  is  the  distance  (length  of  path)  it  would 
travel  in  a  unit  of  time  if  the  velocity  it  had  at  the  instant  con- 
sidered were  uniform.  For  instance,  suppose  the  speed  of  a  body 
were  such  that  at  the  instant  it  arrived  at  a  certain  fixed  point,  it 
would  travel  852  feet  in  one  minute  if  it  continued  at  the  same 
rate  of  speed  for  one  minute;  then  the  velocity  at  the  point 

852 
considered  is  852  feet  per  minute,  or  -^r  =  14.2  feet  per  second. 

If  the  velocity  in  miles  per  hour  were  desired,  note  that  if  the  body 
kept  up  the  same  rate  of  speed  for  one  hour  =  60  minutes,  it 

would  travel   852  X  60  =  51,120   feet  =  -=KH7f  =  9.6TV   miles; 

hence,  the  velocity  at  the  instant  considered  is  9.6tt  miles  per 
hour. 

20.  Acceleration. — Acceleration  is  rate  of  change  of  velocity. 
For  example,  if  the  velocity  of  a  body  is  not  uniform,  it  either 
increases  or  decreases,  and  the  rate  of  increase  or  decrease  is 
called  acceleration,  when  the  rate  of  change  is  uniform,  the  accel- 
eration is  also  uniform.  Thus,  suppose  the  velocity  of  a  body  at  a 
certain  instant  were  120  feet  per  second,  and  4  seconds  later  the 
velocity  were  180  feet  per  second.  If  the  gain  in  speed  were 
constant,  the  acceleration  would  also  be  constant,  and  the  gain 

in  velocity  would  be  at  the  rate  of  -    —r — —  =    15   feet  per 

second  every  second;  this   would  be  expressed  as  15  feet  per 


§4  MATTER,  MOTION,  AND  FORCE  9 

second  per  second,  or  15  ft.  per  sec.2,  which  latter  form  means 
that  the  time  in  seconds  is  to  be  squared.  The  unit  of  accelera- 
tion is  almost  invariably  taken  as  one  foot  per  second  per  second 
=  1  ft.  per  sec.2  or  as  one  centimeter  per  second  per  second  =  1 
cm.  per  sec.2 

If,  in  the  case  just  mentioned,  the  velocity  had  decreased 

uniformly  from  120  ft.  per  sec.  to  80  ft.  per  sec,  the  acceleration 

cq 1 20 

would  have  been  r =  — 10  ft.  per  sec.2,  the  acceleration  in 

this  case  being  negative. 

To  understand  clearly  the  reason  for  squaring  the  time  in  the 
unit    of    acceleration,   perform    the    calculation    by  using  the 

SGC  SGC 

units  in  connection  with  the  numbers.     Thus, '-. ' 

4  sec. 

(8°  -  120)  ^  ft 

= -a — "  =  ~  10~,  =  -  10ft.  per.  sec.2 

4  sec.  sec.2 


FORCE,  MASS,  AND  WEIGHT 

21.  Force. — By  reason  of  its  inertia,  a  body  cannot  put  itself 
in  motion  or  bring  itself  to  rest  or  change  its  shape;  to  do  any  of 
these  things,  the  body  must  be  acted  upon  by  some  force.  There- 
fore, force  may  be  defined  as  that  which  tends  to  change  the  state 
of  rest  or  motion  of  a  body  or  which  acts  to  produce  deformation 
(change  of  shape)  of  a  body.  In  the  case  of  a  gas,  the  force  due 
to  the  motion  of  the  molecules  causes  the  body  to  expand  and 
fill  the  vessel  that  contains  it;  in  the  case  of  a  liquid,  the  force 
due  to  the  earth's  attraction  causes  the  liquid  to  spread  and  con- 
form to  the  shape  of  the  vessel  holding  it,  and  makes  the  upper 
surface  flat. 

Forces  are  called  by  various  names,  according  to  the  effects 
they  produce,  some  of  them  being:  attraction,  repulsion,  adhesion; 
cohesion,  gravitation  (or  weight),  action,  reaction,  friction,  etc. 
But,  whatever  the  effect  produced,  every  force  is  equivalent  to 
a  pull  or  a  push.  A  pull  usually  tends  to  elongate  (stretch)  the 
body  on  which  it  acts,  and  a  push  usually  tends  to  compress 
(shorten)  the  body  on  which  it  acts. 

22.  Cohesion  and  Adhesion. — Cohesion  is  the  force  that  holds 
molecules  together  to  form  a  body;  it  is  an  attractive  force 


10 


ELEMENTS  OF  PHYSICS 


§4 


between  like  molecules.  The  force  of  cohesion  is  extremely 
weak  in  the  case  of  gases  and  extremely  strong  in  the  case  of 
solids.  When  two  bodies  are  held  together  by  the  attraction  of 
unlike  molecules,  the  attracting  force  is  called  adhesion;  thus, 
two  pieces  of  wood  may  be  held  together  by  glue,  and  the  mole- 
cules of  glue  being  different  from  those  of  wood,  the  force  holding 
the  bodies  together  is  called  adhesion.  Likewise  grease  adheres  to 
iron,  wax  to  paper,  etc. 

23.  Measure  of  Force. — -The  force  with  which  the  earth 
attracts  bodies  on  or  near  its  surface  is  called  the  attraction 
(or  force )  of  gravity  or,  more  simply,  gravity  or  weight.  Weight, 
then,  is  the  force  (pressure  or  push)  exerted  by  a  body  resting 
on  the  earth  or  on  another  body.  If  the  body  be  suspended  from 
another  body,  the  connection  being  a  flexible  bodj^,  say  a  string 
or  a  spring,  it  will  exert  a  pull  on  the  second  body  and  will 
stretch  the  string  or  spring.  In  either  case,  the  effect  of  the 
force  will  be  exactly  the  same.  The  force  of  gravity 
is,  therefore,  very  convenient  for  comparing  forces. 

The  force  of  gravity  always  acts  in  the  direction 
of  a  line  drawn  from  the  center  of  gravity  of  the  body 
to  the  center  of  the  earth;  this  line  is  called  a  vertical 
line,  and  any  line  perpendicular  to  it  is  a  horizontal 
line.  It  will  thus  be  seen  that  any  line  (imaginary) 
drawn  on  the  upper  (free)  surface  of  a  liquid  is  a 
horizontal  line.  A  vertical  line  is  also  called  a  plumb 
line,  because  if  a  plumb  bob  be  suspended  from  a 
string,  the  centerline  of  the  string  will  point  to  the 
earth's  center  when  the  plumb  comes  to  rest;  hence, 
this  line  is  vertical. 

24.  Fig.  1  shows  a  spring  balance,  from  the  hook 
of  which  is  suspended  a  standard  one-pound  weight; 
the  indicator  then  points  to  the  1  mark  on  the 
scale.  If,  now,  the  one-pound  weight  be  replaced 
by  another  body  that  brings  the  indicator  to  the  1 
mark,  the  force  exerted  in  both  cases  is  the  same.  Con- 
sequently, the  force  with  which  either  body  presses  against 
any  other  body  on  which  it  rests  is  called  one  pound.  It  is 
evident  that  a  weight  of  one  pound  will  exert  a  force  (pressure)  of 
one  pound  when  resting  on  another  body;  consequently,  forces 
are  measured  by  their  equiva'ent  in  weights,  and  are  expressed  in 


*\ 

0 

!=—  o 

fc 

^—3 

B-4 

X 

1— « 

;: 

|j-7 

fcT* 

P~» 

i^—10 

I          o 

— - 

»**^ 

Fig.  1. 


§4  MATTER,  MOTION,  AND  FORCE  11 

pounds,  grams,  or  kilograms.  For  instance,  the  force  exerted  by 
a  hammer  in  driving  a  nail  a  certain  distance  into  a  board  would 
be  measured  by  the  weight  of  a  body  that,  resting  on  the  nail, 
would  press  it  the  same  distance  into  the  wood  in  the  same  time; 
the  force  with  which  steam  presses  against  a  piston  of  an  engine 
is  equal  to  the  weight  that  would  exert  an  equal  pressure  on 
the  piston.  Observe  that  both  weight  and  force  are  measured 
with  a  spring  balance,  the  measure  of  either  being  determined 
by  the  amount  that  the  spring  is  stretched,  as  indicated  by 
the  scale. 

25.  Mass. — By  mass  is  meant  the  amount  of  matter  contained 
in  a  body.  A  little  consideration  will  show  that  the  mass  of  a 
body  cannot  be  determined  by  measuring  its  volume.  A  cubic 
foot  of  spruce  wood  will  weigh,  say,  35  pounds,  while  a  cubic  foot 
of  cast  iron  will  weigh  450  pounds;  it  is  evident,  therefore,  that  a 
cubic  foot  of  cast  iron  contains  more  matter  than  a  cubic  foot  of 
spruce.  In  other  words,  the  mass  of  cast  iron  is  almost  13  times 
as  great  as  the  mass  of  an  equal  volume  of  spruce  wood.  Conse- 
quently, the  mass  of  a  body  is  determined  by  weighing  it;  but 
the  mass  is  not  equal  to  the  weight,  as  will  now  be  shown. 

26.  If  a  body  be  allowed  to  fall  freely  in  a  vacuum  at  a  place 
whose  latitude  is  about  41°,  or  very  nearly  that  of  New  York,  the 
uniform  acceleration  will  be  32.16ft.  persec.2  The  body  will  have 
this  acceleration  no  matter  what  its  weight,  whether  it  be  1  ounce 
or  1  ton,  because  it  makes  no  difference  whether  the  body  be  di- 
vided into  2000  X  16  =  32000  equal  parts  and  all  fall  separately 
or  whether  they  be  joined  together  and  all  fall  as  one  body.  The 
force  which  causes  the  body  to  fall  and  gives  it  the  acceleration  it 
receives  is  the  force  of  gravity,  which  is  equal  to  the  weight  of  the 
body,  and  this  is  a  constant  (uniform)  force  acting  on  the  body 
during  its  time  of  falling.  It  is  therefore  plain  that  if  a  body 
weighing  one  pound  receive  an  acceleration  of  32.16  ft.  per  sec.2 
when  falling  freely,  a  force  of  one  pound  acting  on  a  body  weighing 
one  pound  and  free  to  move  will  give  it  the  same  acceleration.  It 
would  also  seem  as  though  a  force  of  one  pound  acting  on  a  body 
weighing  2,  3,  w  times  as  much,  that  is,  2,  3,  n  pounds,  would 

receive  an  acceleration  of  only  §,  ^d,  -  th  as  much,  and  experi- 
ment shows  this  to  be  the  case.  Therefore,  if  /  =  the  force 
acting  on  a  body,  w  =  the  weight  of  the  body,  a  =  the  accelera- 


12  ELEMENTS  OF  PHYSICS  §4 

tion  produced  by  the  action  of  the  force  /,  and  g  =  the  accelera- 
tion produced  by  gravity, 

/  :  w  =  a  :  g 

iva       w  k 
f=-==Xa  =  ma, 
9         Q 
w 
when  m  =  —     To  find  the  value  of  w,  it  must  be  found  by  weigh- 
ing the  body  with  a  beam  scale  instead  of  a  spring  balance  (in 
order  that  standard  weights  may  be  used;  the  spring  balance  gives 
true  weight  only  under  specific  conditions) ;  the  value  of  g  may 
be  found  by  direct  experiment  or  by  calculation;  then,  knowing 
the  value  of  g  for  any  locality  and  weighing  the  body  under 
consideration  on  a  beam  scale,  the  value  of  m  can  be  found 
by  substituting  the  known  values  in  the  formula.     The  value 
thus  found  for  m  is  called  the  mass  of  the  body.     From  the  pre- 
ceding equation,  /  =  ma,  that  is,  the  force  which  will  give  to  any 
body  an  acceleration  a  is  equal  to  the  ?nass  of  the  body  multiplied  by 
the  acceleration. 

27.  The  reason  for  defining  the  mass  of  a  body  as  its  weight  at 
any  particular  place  divided  by  the  value  of  g  at  that  place  is 
that  the  value  of  g  is  different  for  different  places  on  the  earth's 
surface;  it  is  least  at  the  equator  and  greatest  at  the  poles,  be- 
cause the  earth  is  not  a  perfect  sphere,  a  point  on  the  surface 
at  the  poles  being  nearer  the  earth's  center  than  a  point  on  the 
surface  at  the  equator.  The  farther  the  body  is  from  the  center 
the  smaller  is  the  value  of  g;  consequently,  the  value  of  g  is  less 
at  the  top  of  a  high  mountain  than  at  sea  level.  The  weight 
of  a  body  varies  directly  as  the  value  of  g,  but  the  mass,  which 
measures  the  quantity  of  matter  remains  constant.  Letting  w' 
and  w"  (read  w  prime  and  w  second)  be  the  weights  of  a  body  at 
two  different  places,  m  the  mass  of  the  body,  and  g'  and  g"  the 
values  of  the  acceleration  produced  by  gravity  at  those  places, 

w'  :  w"  =  g'  :  g" 

w'       w" 
or  w'g"  =  w"g',  from  which  —  =  — 77  =  w.     This  last  equation 

shows  that  m  is  constant,  since  if  g  decreases,  w  also  decreases, 
and  if  g  increases  w  also  increases,  both  in  the  same  proportion. 

28.  The  C.  G.  S.  System.— In  what  is  called  the  C.  G.  S. 
system  (C.  G.S.  is  the  abbreviation  for  centimeter-gram-second), 
the  value  of  g  is  expressed  in  centimeters  per  sec2.     For  latitude 


§4  MATTER,  MOTION,  AND  FORCE  13 

45°,  which  is  half  way  between  the  equator  and  the  poles, 
g  =  980.665  cm.  per  sec.2,  very  nearly,  and  this  value  of  g  is 
now  quite  generally  accepted  as  the  standard  by  scientists.  Then, 
since  1  cm.  =  .03280843  ft.,  980.665  cm.  =  980.665  X  .03280843 
=  32.1741  ft.  For  latitude  of  New  York,  g  =  980.223  cm.  per 
sec.2  =  32.1599,  say  32.16  ft.  per  sec2. 

29.  Weight. — As  previously  stated,  weight  is  caused  by  the 
attraction  of  the  earth.     This  attraction  is  mutual,  the  body 
attracting  the  earth  as  much  as  the  earth  attracts  the  body.     The 
mass  of  the  earth  is  so  great,  however,  that  the  distance  the  earth 
moves  is  too  small  to  be  measured.     The  equatorial  diameter  of 
the  earth  is  about  26.4  miles  greater  than  the  polar  diameter; 
in  other  words,  a  person  standing  at  the  pole  (north  or  south  pole) 
is  over  13  miles  nearer  the  center  than  when  he  stands  on  the 
equator.     The  nearer  a  body  is  to  the  center  the  greater  is  the 
attractive  force  (force  of  gravity),  the  greater  is  the  value  of  g, 
and  the  greater  is  the  weight  of  the  body  when  measured  with 
a  spring  balance.     If,  however,  a  body  is  weighed  with  a  beam 
scale,  the  weight  will  be  the  same  anywhere,  because  the  weight  of 
the  counterpoise  changes  in  the  same  proportion  as  the  weight  of 
the  body.     For  example,  if  a  barrel  of  sugar  weighs  300  pounds  at 
New  Orleans  on  a  beam  scale,  it  will  weigh  the  same  at  New  York, 
at  London,  or  at  the  pole,  using  the  same  scale.     But  if  a  spring 
balance  be  used  that  has  been  graduated  in  accordance  with  the 
value  of  g  =  980.665  cm.  per  sec.2  =  32.1741  ft.  per  sec.2,  the 
weight  recorded  at  New  Orleans  will  not  be  the  true  weight. 
The  value  of  g  for  New  Orleans  is  32.1303  ft.  per  sec.2     The 
true  weight  is. the  weight  that  would  be  recorded  at  a  place  where 
the  value  of  g  is  32.1741  ft.  per  sec.2,  and  may  be  found  from  the 
proportion  of  Art.  27,  or  300  :  w  =  32.1303  :  32.1741,  from  which 
w  =  300.409  pounds.     This  is  the  weight  that  would  be  recorded 
on  a  beam  scale,  if  the  weight  of  the  counterpoises  were  stand- 
ardized to  correspond  with  g  =  980.665  cm.  per  sec.2  =  32.1741 
ft.  per  sec.2 

In  commercial  transactions  and  in  ordinary  engineering  calcu- 
lations, no  attention  is  paid  to  the  variation  in  the  value  of  force 
(or  weight)  due  to  difference  in  latitude;  but  in  all  scientific 
investigations  requiring  accuracy  in  results,  these  variations 
must  be  considered.  For  all  practical  purposes,  the  value  of  any 
force  at  any  place  may  be  considered  as  the  equivalent  of  the 
weight  that  will  produce  the  same  effect  at  that  place. 


14  ELEMENTS  OF  PHYSICS  §4 

30.  Density. — Density  is  the  quantity  of  matter  contained  in  a 
unit  of  volume;  it  is  the  tnass  of  a  unit  of  volume.  If  the  density 
of  a  body  be  represented  by  D,  the  mass  by  m,  the  volume  by  V, 
the  weight  by  w,  and  the  acceleration  due  to  gravity  by  g, 

m       _^ 

V       gV  {i) 

Solving  this  equation  for  m, 

m  =  DV  (2) 

that  is,  the  mass  of  any  body  is  equal  to  the  product  of  its  density 
and  volume. 

Density  is  also  frequently  denned  as  the  weight  of  a  unit  of 
volume,  in  which  case, 

D  =  'f  (3) 

The  density  of  gases  is  almost  always  expressed  as  in  the  second 
of  the  above  definitions.  If  the  unit  of  weight  be  taken  as  1 
pound  and  the  unit  of  volume  as  1  cubic  foot,  the  density  of  a 
body  by  the  second  definition  is  in  pounds  per  cubic  foot,  and  the 
density  may  always  be  found  by  weighing  1  cubic  foot.  Another 
name  for  the  weight  of  a  unit  of  volume  is  specific  weight;  thus, 
if  the  weight  of  a  cubic  foot  of  cast  iron  is  450  pounds,  its  specific 
weight  is  450  pounds  per  cubic  foot,  which  is  also  the  density  in 
accordance   with   the   second   definition.     In   accordance   with 

450 
the    first    definition,    the    density    of    cast    iron  is  ™y-«ttty 

J  32.1/41X1 

13.98  — .  There  is  no  name  for  the  unit  of  density  as  determined 
by  formula  (1),  which  may  be  called  the  specific  mass;  neither  is 
there  any  for  the  unit  of  mass.  Consequently,  for  the  want  of  a 
better  name,  these  units  may  be  called  density  units  and  mass 
units,  respectively. 

WORK  AND  ENERGY 

31.  Work. — That  point  of  a  body  at  which  a  force  is  applied  or 
at  which  it  may  be  considered  as  acting  is  called  the  point  of 
application.  If.  as  the  result  of  the  action  of  a  force,  the  point  of 
application  is  moved  through  a  certain  distance,  work  is  dour. 
Thus,  if  a  body  weighing  15  pounds  is  lifted  6  feet  vertically,  work 
is  done  in  doing  this.  The  unit  of  work  is  usually  taken  as  afoot- 
pound;  this  is  a  compound  unit,  and  means  that  a  resistance  of  one 
pound  has  been  overcome  through  a  distance  of  one  foot,  and  is 


§4  MATTER,  MOTION,  AND  FORCE  15 

exactly  equivalent  to  1  pound  X  1  ft.  In  other  words,  to  find  the 
work  done,  multiply  the  force  (in  pounds)  by  the  distance  (in  feet) 
through  which  it  acts.  The  work  done  in  raising  15  pounds  through 
a  vertical  distance  of  6  ft.  is  15  X  6  =  90  ft. -lb. 

Suppose  the  average  pressure  on  the  piston  of  a  steam  engine  is 
63  pounds  per  square  inch,  that  the  diameter  of  the  piston  is 
15  inches,  and  that  it  moves  18  inches  during  one  stroke;  then  the 
work  done  per  stroke  may  be  found  as  follows.  The  total  average 
pressure  during  the  stroke  is  63  times  the  area  of  the  piston  in 
square  inches,  since  the  average  pressure  is  63  pounds  per  sq.  in.; 

7T 

hence,  total  pressure  =  63  X  t  X  152.     The  distance  through 

18 
which  this  pressure  (force)  acts  is  18  in.  =  tt,  =  1.5  ft.     Conse- 
quently,   the   work   done   is    1.5  X  63  X  .7854  X  152  =  16,700 
ft.-lb. 

32.  Work  is  the  overcoming  of  a  resistance  through  a  distance;  if  a 
force  acts  on  a  body  without  moving  the  point  of  application,  no 
work  is  done.  For  instance,  if  a  man  try  to  lift  a  stone  that  is  too 
heavy  for  him  to  move,  he  will  apply  considerable  force  to  the 
stone,  but  no  work  will  be  done  on  the  stone.  In  the  case  of  a 
falling  body,  gravity  is  the  acting  force,  and  equals  the  weight  of 
the  body;  the  work  done  is  the  weight  of  the  body  multiplied  by 
the  vertical  distance  through  which  the  body  falls. 

Observe  that  the  work  done  is  not  dependent  on  the  time  it 
takes  to  do  it.  Thus,  15  pounds  raised  through  a  vertical  dis- 
tance of  6  feet  is  equal  to  15  X  6  =  90  ft.-lb.  of  work;  and  it 
makes  no  difference  whether  it  took  one  second  to  raise  the 
weight  or  one  month,  the  work  done  is  the  product  of  the  acting 
force  and  the  distance  through  which  it  acts.  It  should  also  be 
noted  that  when  gravity  is  the  acting  force,  it  makes  no  difference 
how  the  body  gets  from  one  level  to  the  other  (that  is,  what  the 
shape  of  the  curve  representing  its  path),  the  work  done  is  the 
weight  of  the  body  multiplied  by  the  vertical  distance  (perpendic- 
ular distance)  between  the  level  that  includes  the  starting  point 
and  the  level  that  includes  the  stopping  point.  Thus,  in  hauling 
a  wagon  up  a  hill,  suppose  the  weight  of  the  wagon  and  its  load 
is  900  pounds  and  that  the  difference  of  level  between  the  top  of 
the  hill  and  the  bottom  is  120  feet;  then,  neglecting  friction  and 
other  resistances,  the  work  done  is  900  X  120  =  108,000  ft.-lb., 
regardless  of  how  the  top  of  the  hill  was  reached.  The  result  is 
the  same  as  though  the  wagon  had  been  lifted  bodily  120  ft., 


16  ELEMENTS  OF  PHYSICS  §4 

as  by  an  elevator.  This  is  because  when  the  wagon  moves  up  the 
hill,  the  acting  force  is  not  equal  to  the  weight  of  the  body,  but 
is  considerably  less,  depending  on  the  slope.  When  the  body 
moves  vertically  up  or  down,  then  the  acting  force  is  equal  to  the 
weight. 

33.  Energy. — Energy  means  capacity  for  doing  work — ability 
to  do  work.  A  body  in  motion  cannot  instantly  be  brought  to 
rest;  to  bring  it  to  rest,  a  force  must  act  through  a  distance,  no 
matter  how  short.  For  instance,  when  a  blow  is  struck  with  a 
hammer,  a  dent  is  made  in  the  body  struck,  and  the  depth  of  this 
dent  is  the  distance  passed  through  by  the  hammer  in  coming  to 
rest.  This  distance  (in  feet)  multiplied  by  the  force  of  the  blow 
is  the  work  done.  At  the  instant  the  hammer  strikes,  but  before 
any  work  is  done,  the  hammer  has  capacity  for  doing  the  work 
that  is  done  as  the  result  of  the  blow,  and  this  capacity  for  doing 
work  is  called  energy. 

Energy  is  measured  in  the  same  units  as  work,  that  is,  in  foot- 
pounds. Suppose  a  body  weighing  500  pounds  to  fall  through  a 
vertical  height  of  16  ft.;  the  work  it  can  do  is  exactly  equal  to  the 
work  required  to  be  done  on  the  body  to  raise  it  through  a  vertical 
height  of  16  ft.,  neglecting  resistance  of  the  air,  friction,  etc.  in 
both  cases.  This  work  is  500  X  16  =  8000  ft. -lb. ;  consequently, 
the  energy  of  the  body  is  8000  ft. -lb.  also. 

34.  Kinds  of  Energy. — It  is  customary  to  divide  energy  into 
two  classes — 'potential  energy  and  kinetic  energy.  Potential 
energy  is  that  due  to  the  position  of  a  body.  Consider,  for 
example,  a  pile  driver.  When  the  hammer  that  drives  the  pile 
has  been  raised  to  its  highest  point  and  is  ready  to  be  released,  it 
possesses  energy  due  to  its  position,  the  amount  being  equal  to  the 
weight  of  the  hammer  in  pounds  multiplied  by  the  height  in  feet 
of  the  hammer  above  the  pile,  which  is  equal  to  the  vertical 
distance  between  the  bottom  of  the  hammer  and  the  top  of  the 
pile.  This  energy  is  potential  energy,  so  called  because  it  may  or 
may  not  be  used.  Potential  means  possible;  hence,  potential 
energy  means  possible  energy. 

Kinetic  energy  is  actual  energy — it  is  the  energy  of  a  body  in 
motion.  In  the  previous  illustration,  when  the  hammer  is 
released,  it  falls  and  the  potential  energy  is  gradually  changed 
into  kinetic  energy  until,  at  the  instant  the  hammer  hits  the  top 
of  the  pile,  all  the  potential  energy  has  been  changed  into  kinetic 


§4  MATTER,  MOTION,  AND  FORCE  17 

energy.  The  sum  of  the  potential  energy  (if  any)  and  the 
kinetic  energy  (if  any)  is  the  total  energy.  Thus,  when  the 
hammer  has  fallen  through  }ith  the  height,  ^ith  of  the  potential 
energy  has  been  changed  to  kinetic  energy;  hence,  the  total 
energy  is  made  up  of  %th  of  the  original  potential  energy  and  the 
other  3^th  is  kinetic  energy.  Again,  when  a  rifle  is  fired,  the 
potential  energy  of  the  powder  is  changed  to  kinetic  energy  in  the 
barrel  of  the  gun ;  the  kinetic  energy  is  converted  into  work  on  the 
bullet,  and  when  the  bullet  leaves  the  gun,  it  has  kinetic  energy, 
which  is  converted  into  work  when  the  bullet  is  stopped.  It  will 
be  noted  that  the  bullet  does  not  have  potential  energy  at  any 
time;  it  has  either  no  energy  at  all  or  it  has  kinetic  energy.  If, 
however,  the  bullet  be  fired  vertically  upward,  and  it  meet  with  no 
resistance  from  the  air,  the  action  of  gravity  causes  the  velocity 
to  decrease,  until  it  stops,  but  immediately  begins  to  fall.  Dur- 
ing the  rise,  the  kinetic  energy  has  been  changing  into  potential 
energy,  and  at  the  instant  it  begins  to  fall,  all  the  kinetic  energy 
has  been  changed  to  potential  energy.  As  it  falls,  the  potential 
energy  decreases  and  the  kinetic  energy  increases;  and  when 
it  again  reaches  the  earth,  all  the  potential  energy  has  been 
changed  to  kinetic  energy,  equal  in  amount  to  the  original  kinetic 
energy  it  had  when  it  left  the  gun. 

All  work  is  the  result  of  kinetic  energy,  and  potential  energy 
of  some  kind  must  be  converted  into  kinetic  energy  before  any 
work  can  be  done.  In  the  case  of  the  gun,  the  potential  energy 
was  in  the  powder;  in  the  case  of  a  steam  engine,  the  potential 
energy  is  in  the  steam;  and  in  the  case  of  the  pile  driver,  the 
potential  energy  was  in  the  position  of  the  hammer  before  it  was 
released.  Potential  energy  is  always  stored  energy,  while  kinetic 
energy  is  active  energy  that  is,  or  can  be,  transformed  into  work. 

Energy  exists  in  innumerable  forms;  like  matter,  it  cannot  be 
destroyed,  but  may  be  changed  in  various  ways.  The  energy 
stored  in  coal  was  received  from  the  sun  millions  of  years  ago; 
it  is  changed  into  heat  (a  form  of  kinetic  energy),  by  combus- 
tion, and  generates  steam;  steam  drives  the  engine,  which,  in 
turn,  drives  a  dynamo  that  generates  electricity;  thus  energy  that 
originally  came  from  the  sun  is  converted  into  electric  energy, 
used  to  drive  motors,  light  lamps,  etc.  The  energy  stored  in 
food  is  converted  into  muscular  energy,  and  enables  us  to  live, 
move,  work,  and  play. 


18  ELEMENTS  OF  PHYSICS  §4 

HYDROSTATICS 


PASCAL'S  LAW 


FLUID  PRESSURE 

35.  Equilibrium. — A  body  is  said  to  be  in  equilibrium  when 
it  is  at  rest  under  the  action  of  forces  or  if  in  motion,  there  is  no 
change  in  its  velocity  or  direction  of  motion.  A  block  of  stone 
resting  on  the  earth  is  in  equilibrium;  the  forces  acting  on  the 
stone  are  the  force  of  gravity,  which  acts  downwards  and  pulls 
the  block  toward  the  center  of  the  earth,  and  the  pressure  of  the 
earth  against  the  block,  called  the  reaction  of  the  earth,  which 
acts  upwards  and  prevents  the  block  from  moving  toward  the 
center  of  the  earth.  A  stick  balanced  on  a  knife  edge  is  in 
equilibrium ;  gravity  tends  to  pull  the  part  of  the  stick  on  one  side 
of  the  knife  edge  one  way  and  the  other  part  the  other  way,  the 
two  balancing  each  other;  but  the  total  pull  toward  the  earth  is 
the  weight  of  the  stick,  and  this  is  resisted  by  the  reaction  of  the 
knife  edge,  which  acts  upward.  A  person  riding  on  a  railway 
train  that  is  moving  in  a  straight  line,  along  level  ground,  at  a 
constant  rate  of  speed  is  in  equilibrium;  the  force  exerted  by  the 
locomotive  in  pulling  the  train  just  balances  that  required  to 
overcome  the  resistance  of  the  air,  friction,  etc. ;  but,  if  the  train 
commences  to  go  around  a  curve  or  up  or  down  a  grade,  the 
person  will  no  longer  be  in  equilibrium — the  direction  of  motion 
has  been  changed;  or,  if  the  speed  of  the  train  is  increased  or 
decreased,  he  will  no  longer  be  in  equilibrium — the  velocity  has 
been  changed.  If  the  change  is  sudden,  he  will  be  thrown  for- 
wards, backwards,  sideways,  up,  or  down,  according  to  the  nature 
of  the  change  in  motion. 

Hydrostatics  is  that  branch  of  science  which  treats  of  fluids  at 
rest  under  the  action  of  forces,  that  is,  of  fluids  in  equilibrium ; 
the  word  is  derived  from  the  Greek,  and  literally  means  water 
at  rest,  hydro  meaning  water  and  statics  meaning  standing  or  at 
rest.  Hydrostatics  is  usually  limited  to  water  and  other  liquids, 
but  may  also  be  applied  to  gases. 

36.  Pressure. — When  a  steady  force  having  the  effect  of  a 
push  acts  on  a  body,  it  is  called  a  pressure.  When  the  force 
acts  very  suddenly  and  for  a  short  time,  it  is  called  an  impulse, 


§4  HYDROSTATICS  10 

or  a  blow;  but,  if  desired,  any  blow  may  be  considered  as  a  pres- 
sure acting  for  a  very  short  time. 

Suppose  a  log  having  a  mean  diameter  of  16  in.,  20  ft.  long, 
with  ends  at  right  angles  to  the  axis,  and  weighing  060  pounds  to 
stand  on  one  end ;  it  will  exert  a  pressure  of  060  pounds  on  the 
surface  that  supports  it,  and  this  pressure  will  be  evenly  distrib- 
uted over  a  surface  whose  area  is  equal  to  the  area  of  the  end  of 
the  log.  Since  the  log  is  not  exactly  round,  the  area  of  the  end 
may  be  considered  as  equal  to  the  area  of  a  circle  whose  diameter 

is  the  same  as  that  of  the  log,  but  taking  tt  equal  to  3  instead  of 

3 
3.1416.     Hence,  the  area  is.X  162  =  102  sq.  in.     The  pres- 

•     i  •     i       060       _ 
sure  exerted  by  the  log  on  each  square  inch  is  then  y^  =  5  pounds 

per  square  inch.  When  a  pressure  is  stated  as  a  force  per  unit  of 
area,  it  is  called  specific  pressure,  and  the  pressure  on  the  entire 
area  is  called  the  total  pressure.  In  the  above  case,  the  total 
pressure  was  060  pounds,  and  the  specific  pressure  was  5  pounds 
per  sq.  in. 

37.  Specific  pressures  are  generally  stated  as  pounds  per  square 
inch  or  pounds  per  square  foot,  but  in  connection  with  fluid 
pressure,  they  may  be  stated  as  feet  of  water,  inches  of  mercury, 
or  atmospheres,  terms  that  will  be  explained  later.  In  the  metric 
system,  specific  pressures  are  usually  expressed  in  kilograms  per 
square  centimeter  or  kilograms  per  square  meter;  meters  of  water 
and  millimeters  of  mercury  are  also  used. 

38.  Total  pressures  may  be  considered  as  distributed  over  an 
area  or  as  acting  on  a  line  passing  through  the  center  of  gravity 
of  the  area  pressed  against.  Thus,  in  the  case  of  the  log  standing 
on  one  end,  the  total  pressure  of  060  pounds  may  be  considered  as 
distributed  over  an  area  equal  to  the  area  of  one  end  or  as  a 
concentrated  force  that  acts  at  the  center  of  gravity  of  the  area. 
Insofar  as  any  movement  of  the  body  under  the  action  of  the 
force  (pressure)  is  concerned,  the  effect  will  be  the  same  in  either 
case. 

39.  Transmission  of  Pressure. — When  a  force  (pressure)  acts  on 
a  solid,  it  is  transmitted  through  the  solid  in  a  line  that  forms  an 
extension  of  the  line  of  action;  thus,  if  one  end  of  a  chisel  be  hit 
with  a  hammer  or  mallet,  the  other  end  of  the  chisel  will  impart 
to  any  body  it  touches  a  blow  of  the  same  force  as  that  which  was 
received  by  the  chisel.     In  other  words,  the  chisel  transmitted  the 


20 


ELEMENTS  OF  PHYSICS 


§4 


blow  it  received  at  one  end  to  the  other  end.  The  blow,  itself, 
measured  as  a  force,  may  be  different  at  the  two  ends,  by  reason 
of  a  difference  in  the  shapes  of  the  ends,  and  the  material  of  the 
body  struck  by  the  chisel  will  probably  be  different  from  that  of 
the  chisel;  but  all  the  energy  received  at  one  end  of  the  chisel  will 
be  transmitted  undiminished  to  the  other  end,  provided  the  blow 
is  not  hard  enough  to  injure  (deform)  the  chisel  in  any  way. 
There  will  be  no  pressure  in  any  direction  except  lengthwise  of 
the  chisel.  The  same  is  true  of  a  weight  resting  on  a  pillar;  the 
pressure  due  to  the  weight  is  transmitted  from  one  end  of  the 
pillar  to  the  other,  and  the  total  pressure  is  the  same  at  either 
end. 

In  the  case  of  fluids,  the  result  is  entirely  different,  because  the 
molecules  of  the  fluid  are  free  to  move  in  any  direction.  When 
pressure  is  applied  to  a  solid  that  is  not  too  pliable,  it  retains  its 

shape  for  all  practical  purposes,  and 
any  movement  of  the  molecules  must 
be  lengthwise  only.  There  is  a  cross- 
wise (lateral)  movement,  but  it  is  so 
slight  that  for  ordinary  pressures,  it 
may  be  neglected,  and  the  lengthwise 
(longitudinal)  movement  may  also  be 
neglected. 

40.  When  a  fluid  is  subjected  to 
pressure,  the  pressure  is  transmitted 
undiminished  in  every  direction.  It 
is  to  be  noted  that  the  word  pressure 
in  this  case  means  specific  pressure. 
Referring  to  Fig.  2,  which  represents 
a  vertical  longitudinal  section  passing 
through  the  axis  of  a  cylindrical 
vessel,  suppose  the  vessel  to  be  filled 
with  a  fluid,  say  water,  and  fitted 
with  a  tight-fitting  piston  P,  which 
rests  on  top  of  the  water.  Suppose  further  that  a  weight  W 
of  480  pounds  is  placed  on  top  of  the  piston,  as  shown.  The 
pressure  on  top  of  the  water  due  to  the  weight  W  will  be  480 
pounds  and  this  will  also  be  the  pressure  on  the  bottom  of  the 
vessel  (neglecting  the  weight  of  the  water),  the  result  being 
the  same  as  though  the  space  occupied  by  the  water  were  a 
round  block  of  wood.     If  the  area  of  the  bottom  pf  the  vessel  is 


480  lbs. 

f    * 

1 

1 "H 

F,                       i 

_J 

1              P 

■ 

■J! 

^>^>Sd-D^^. 

m<_ fe. 

imzj^zmm 

.  -4— p-^ 

t    .       .                                                         =3 

Area  60  sq.  in. 
Fig.  2. 


HYDROSTATICS 


21 


60  sq.  in.,  the  specific  pressure  is  -fifr  =8  pounds  per  sq.  in.,  and 

this  specific  pressure  is  transmitted  in  every  direction,  as  indicated 
by  the  arrows — upward,  downward,  laterally,  and  at  any  angle 
whatever. 

Had  the  vessel  been  filled  with  a  gas  instead  of  water,  the  same 
result  would  have  been  obtained,  the  only  difference  being  that 
water  is  only  very  slightly  compressible  and  the  change  in  volume 
is  not  noticeable;  but  gas  is  so  compressible  that  the  volume 
would  be  very  much  smaller.  For  this  reason,  water  will  be 
adopted  as  the  fluid  under  consideration  in  what  follows,  unless 
otherwise  specified;  but  what  is  true  of  water  in  general  is  also 
true  of  any  other  liquid  and  of  gases. 

41.  Pascal's  Law. — The  law  governing  the  transmission  of 
pressure  in  fluids  confined  in  a  closed  vessel  was  discovered  by 
Pascal,  a  famous  French  scientist 
and  mathematician  (1623-1662), 
and  is  known  as  Pascal's  law;  it 
may  be  stated  as  follows : 

Specific  pressure  exerted  on  any 
fluid  confined  in  a  closed  vessel  is 
transmitted  in  all  directions  and 
acts  on  all  surfaces  touched  by  the 
fluid  in  a  direction  perpendicular 
to  those  surfaces. 

42.  If  the  surface  pressed 
against  by  the  fluid  is  a  curved 
surface,  the  perpendicular  to 
the  surface  at  any  point  is 
called  a  normal,  and  a  normal 
to  a  curve  at  any  point  is  per- 
pendicular to  the  tangent  to 
the  curve  at  that  point.  This 
is  illustrated  in  Fig.  3,  where 
ECFG  represents  a  curved  surface,  and  the  arrows  a,  b,  c,  d, 
etc.  represent  normals  to  this  surface.  Let  n  represent  the 
center  of  gravity  of  the  surface,  and  let  P  represent  a  pressure 
which,  acting  on  the  surface  at  n  in  the  direction  of  the  normal 
N  will  produce  the  same  effect  on  the  surface  as  is  produced 
by  the  pressure  of  the  fluid.  To  find  the  value  of  P,  let  RS  be  a 
plane  perpendicular  to  the  normal  N,  and  let  A'B'C'D'  be  the 


Fig.  3. 


22  ELEMENTS  OF  PHYSICS  §4 

projection  of  ECFG  upon  the  plane  RS.  Let  a  =  the  area  of 
A'B'C'D'  and  p  =  the  specific  pressure  exerted  on  the  fluid;  then 

P  =  pa, 

that  is,  the  total  normal  pressure  against  any  surface  is  equal  to 
the  specific  pressure  multiplied  by  the  projected  area  of  the  sur- 
face, projected  on  a  plane  at  right  angles  to  the  normal  drawn 
through  the  center  of  gravity  of  the  surface.  This  total  normal 
pressure  tends  to  move  the  entire  surface  under  consideration  in 
the  direction  of  the  normal.     In  the  case  of  the  cylinder,  Fig. 

2,  the  diameter  of  the  cylinder  is  d  =  A/—  =  \  hr—n— t.    ~  8.74 

\  7r         \3.1416 

in.,  nearly.  Assuming  that  the  depth  of  the  water  in  the  cylin- 
der is  25  in.,  the  projected  area  of  one-half  the  cylinder  on  a  plane 
perpendicular  to  the  normal  through  the  center  of  gravity  of  the 
half-cylinder  (which  will  be  on  a  line  midway  between  the  top 
and  bottom  and  midway  between  the  parallel  sides  of  the  half- 
cylinder)  will  have  the  shape  of  a  rectangle  whose  length  is 
25  in.  and  breadth  is  8.74  in.  The  projected  area  will  then  be 
25  X  8.74  =  218.5  sq.  in.  Since  the  specific  pressure  is  8  lb. 
per  sq.  in.,  the  total  normal  pressure  on  the  half-cylinder  is 
218.5  X  8  =  1748  pounds.  This  same  pressure  acts  on  the  other 
half  of  the  cylinder,  but  in  the  opposite  direction.  Therefore, 
the  total  lateral  pressure  due  to  the  weight  of  480  pounds  tending 
to  separate  one-half  of  the  cylinder  from  the  other  half  is  1748 
pounds  or  almost  four  times  the  weight,  in  this  case. 

43.  Figure  4  represents  a  flask  filled  to  the  line  mn  with  water, 
the  space  between  mn  and  the  bottom  of  the  plunger  p  being 
filled  with  air.  As  the  handle  H  is  pushed  down,  the  air  is  com- 
pressed, being  confined  between  the  top  of  the  water  and  the 
bottom  of  the  plunger,  and  the  greater  the  pressure  the  smaller 
becomes  the  volume  of  the  air.  The  pressure  is  transmitted  in 
all  directions  through  the  air,  from  the  air  to  the  water,  and  in 
all  directions  through  the  water.  Assuming  that  the  pressure 
on  H  is  such  that  the  pressure  on  the  confined  air  is  12  pounds 
per  square  inch,  this  specific  pressure  will  be  transmitted  to  all  sur- 
faces touched  by  the  water.  Little  pistons  are  placed  in  the 
vessel  at  a,  b,  c,  etc.,  and  their  cross-sections  have  the  following 
areas:  a  =  1.5  sq.  in.,  b  =  2.75  sq.  in.,  c  =  2.25  sq.  in.,  d  =  3.5 
sq.  in.,  e  =  4.75  sq.  in.,  /  =  2.5  sq.  in.,  g  =  5.5  sq.  in.,  and 
h  =  3  sq.  in.     Then,  neglecting  the  weight  of  the  water,  the 


HYDROSTATICS 


23 


forces  acting  on  the  inside  faces  of  the  pistons,  tending  to  force 
them  out,  and  which  must  be  resisted  by  an  equal  and  opposite 
force  on  the  other  side  of 
the  pistons,  indicated  by 
the  arrows,  are  respectively 
1.5  X  12  =  18  pounds,  2.75 
X  12  =  33  pounds,  2.25  X 
12  =  27  pounds,  3.5  X  12 
=  42  pounds,  4.75  X  12  = 
57  pounds,  2.5  X  12  =  30 
pounds,  5.5  X  12  =  66 
pounds,  and  3  X  12  =  36 
pounds.  These  are  all 
normal  pressures;  the  pis- 
tons are  supposed  to  be 
round,  but  the  faces  may 
be  flat  or  curved,  the  areas 
given  above  being  the  pro- 
jected areas,  and  equal  to 
the  squares  of  the  diame- 
ters of  the  pistons  multi- 
plied by  .7854. 

44.  Hydrostatic  Ma- 
chines.^— A  hydrostatic 
machine  is  a  device  for 
raising  heavy  loads  or  exerting  great  pressure  as  in  pressing 
bales  of  pulp.  The  principle  governing  their  operation  is 
illustrated  in  Fig.  5.     P  and  p  are  pistons  working  in  the  cyl- 


Fig.  5. 


inders  A  and  B.     P  carries  a  weight  represented  by  II*,  and  the 
force  pushing  p  down  is  represented  by  the  weight  w.     The  space 


24  ELEMENTS  OF  PHYSICS  §4 

beneath  the  pistons  is  filled  with  water,  oil,  or  other  liquid,  and 
the  two  cylinders  are  connected  by  a  pipe  a,  so  that  the  liquid 
can  flow  freely  from  one  cylinder  to  the  other.  As  p  moves  down 
the  liquid  in  J.  flows  through  a  into  B,  causing  P  to  rise  and  with 
it  the  weight  W.  The  specific  pressure  exerted  by  p  is  trans- 
mitted to  P.  Let  d  =  diameter  of  p  and  D  =  diameter  of  P; 
then  .7854  d2  =  area  of  p  and  .7854  D2  =  area  of  P,  and  the  speci- 

•  .        w 
fie  pressure  exerted  by  w  is  „„-,  ,2.     But  this  must  be  the  same 

as  the  specific  pressure  exerted  on  W,  which  must  be     7oe4.n2 

therefore,   Tg^2  =  j^W2'  0r  I  =  D*>  fr°m  which 

w  d2 
That  is,  the  weight  that  can  be  lifted  (or  the  pressure  that 
can  be  exerted)  by  the  large  piston  is  equal  to  the  product  of 
the  weight  (or  force  exerted)  on  the  small  piston  and  the  square 
of  the  diameter  of  the  large  piston  divided  by  the  square  of  the 
diameter  of  the  small  piston. 

Example. — If  the  diameter  of  the  small  piston  in  Fig.  5  is  13^  in.,  of  the 
large  piston  12  in.,  and  the  force  exerted  on  the  small  piston  is  38  lb.,  what 
weight  W  can  be  raised  by  the  large  piston? 

Solution. — Substituting  in  the  above  formula  the  values  given, 

W  =  — =-=z —  =  2432  pounds.     Ans. 
1.5^ 

45.  Attention  is  called  to  the  fact  that  the  work  done  on  the 
small  piston  is  (neglecting  friction  and  other  hurtful  resistances) 
exactly  the  same  as  the  work  done  on  the  large  piston.  Let  h  = 
the  distance  (height)  passed  through  by  the  small  piston  and 
H  =  the  distance  passed  through  by  the  large  piston ;  then  wh  = 

WH,  from  which  H  =  ^r-     If,  in  the  example  just  given,  w 

38  X  2 
moves  2  in.,  TT'  will  move  only    04.00    =  -03125  in.     In  other 

words,  what  is  gained  in  pressure  or  lifting  force  is  lost  in  speed. 
This  is  a  general  law,  and  is  true  of  any  machine.  The  converse 
is  equally  true;  that  is,  if  there  is  a  gain  in  speed,  there  is  a  loss 
in  the  force  that  can  be  applied.  For  example,  suppose  that, 
referring  to  Fig.  5,  it  were  desired  to  move  the  small  piston  upward 
6  in.  while  the  large  piston  was  moving  downward  ^  in.;  if  the 
force  W  exerted  on  the  large  piston  is  4000  pounds,  what  pres- 


§4  HYDROSTATICS  25 

sure  will  be  exerted  on  the  small  piston?     Since  wh  =  WH, 

wX  6  =  4000  X  .5  =  2000,    and    w  =  -g-  =  333£    pounds. 

In  other  words,  a  force  of  4000  pounds  moving  }i  inch  can  raise 
a  weight  of  only  333^  pounds  through  a  height  of  6  inches,  and 
what  is  gained  in  speed  (distance)  is  lost  in  applied  force. 

46.  The  conclusion  arrived  at  in  the  last  article  may  also  be 
verified  as  follows:  Taking  the  dimensions  of  the  example  in 
Art.  44,  when  the  small  piston  moves  down,  it  forces  an  amount 
of  liquid  into  the  cylinder  B  that  is  equal  to  the  area  of  the  small 
piston  multiplied  by  the  distance  through  which  it  moves;  that 
is,  it  is  equal  to  the  volume  of  a  cylinder  whose  diameter  is  the 
same  as  that  of  the  piston  and  whose  altitude  is  equal  to  the 
distance  that  the  piston  moves.  The  volume  of  the  liquid  under 
the  large  piston  is  increased  the  same  amount,  and  is  equivalent 
to  a  cylinder  whose  diameter  is  that  of  the  large  piston  and 
whose  altitude  is  the  distance  the  piston  is  raised,  or  H.  Con- 
sequently, assuming  that  piston  p  moves  2  inches,  .7854  X  1.52 

X  2  =  .7854  X  122  X  H,  or  H  =  ^^  2  =  ™  in.  =   .03125 

in.,  and  what  is  gained  in  applied  force  is  lost  in  speed.  Observe 
that  this  result  is  exactly  the  same  as  that  obtained  in  Art.  45. 

47.  Pressure  of  Liquid  Due  to  its  Own  Weight. — In  what  has 
preceded,  only  the  effects  produced  by  the  application  of 
an  external  pressure  have  been  considered.  Since  liquids  have 
weight,  they  exert  pressure  on  the  vessels  containing  them,  and 
the  methods  of  determining  this  pressure  will  now  be  considered. 
Referring  to  Fig.  2,  suppose  the  piston  and  its  weight  of  480 
pounds  to  be  removed;  the  total  pressure  on  the  bottom  of  the 
cylinder  will  then  evidently  equal  the  weight  of  the  liquid,  and 
the  specific  pressure  on  the  bottom  will  equal  the  weight  of  the 
liquid  divided  by  the  area  of  the  bottom,  and  will  be  transmitted 
downward,  upward,  and  laterally.  This  specific  pressure  will 
not,  however,  be  the  same  for  all  parts  of  the  liquid,  since  the 
pressure  on  the  top  of  the  liquid  will  be  0.  At  any  point  be- 
tween the  top  of  the  liquid  and  the  bottom  of  the  vessel,  the 
pressure  will  be  equal  to  the  weight  of  the  liquid  corresponding 
to  the  depth  of  the  point  below  the  top  of  the  liquid.  Insofar  as 
the  downward  pressure  is  concerned,  the  case  is  exactly  analo- 
gous to  a  pile  of  bricks,  see  Fig.  6.  The  pressure  on  the  surface 
AB  that  supports  the  bricks  is  equal  to  the  weight  of  all  the 


2G 


ELEMENTS  OF  PHYSICS 


§4 


Fig.  6. 


bricks,  or  7  bricks  in  this  instance.  There  are  6  bricks  on  top 
of  brick  No.  1,  and  the  pressure  on  the  top  of  this  brick  is 
equal  to  the  weight  of  6  bricks,  the  depth  of  the  pile  between 
brick  1  and  the  top  of  brick  7.  The  pressure  on  top  of  brick  4 
is  3  bricks,  the  number  of  bricks  between  the  top  of  brick  4 
and  the  top  of  brick  7;  etc. 

Suppose  the  size  of  the  bricks  to  be  4  in.  wide,  8  in.  long,  and 
2  in.  thick,  and  that  each  brick  weighs  4.5  pounds.     The  total 

pressure  on  the  surface  AB  is  7X  4.5 
=  31.5  pounds,  and  the  specific 
pressure  is  31.5  -*-  4x8  =  .984375 
pound  per  square  inch.  The  weight 
of  1  cubic  inch  of  brick  is  4.5  -j-  (4  X 
8X2)=  .0703125  pound.  The  depth 
;  of  the  pile  is  7  X  2  =  14  in.,  and 
.0703125  X  14  =  .984375  pound  =  the 
weight  of  a  prism  of  brick  1  in.  square 
and  14  in.  high.  But  this  value  is 
the  same  as  the  specific  pressure; 
hence,  the  specific  pressure  for  any 
depth  is  equal  to  the  weight  of  a  prism  of  the  brick  whose  base 
is  the  unit  of  area  and  whose  height  is  the  depth  of  the  brick 
at  the  point  considered.  '  This  also  applies  to  fluids. 

48.  A  cube  of  water  measuring  1  ft.  on  each  edge  (1  cubic  foot) 
weighs  62.4  lb.  at  its  temperature  of  maximum  density  (4°C.  or 
39.2°F.);  it  weighs  less  at  higher  temperatures,  but  unless  very 
exact  results  are  desired,  the  weight  of  water  may  be  taken  as 
62.4  pounds  per  cu.  ft.  A  column  of  water  1  in.  square  and  1  ft. 
high  evidently  weighs  62.4  -J-  144  =  H  =  .4}  £  pound.  There- 
fore, for  water  at  any  depth,  let  p  =  the  specific  pressure  and 
h  =  the  depth  in  feet;  then, 

p  =  Alh  =  ||A  (1) 

If  the  depth  be  taken  in  inches,  the  weight  of  1  cu.  in.  =  62.4 

4-  1728  =  ^6a0  =  .036111  pound,  say  .0361  pound  for  practical 

purposes.     Letting  h'  be  the  depth  in  inches,  the  specific  pressure  is 

V  =  iM'  =  -0361//  (2) 

If  it  be  desired  to  find  the  depth  necessary  to  produce  a  given 

specific  pressure,  solve  the  above  formulas  for  h  and  h' ,  obtaining 

h  =  UV  =  2.3077p  (3) 

and 


h'  = 


Wp 


27.692p 


(4) 


§4 


HYDROSTATICS 


27 


Fig.  7. 


For  practical  purposes,  a  depth  of  2.31  ft.  or  27.7  in.  of  water 
may  be  considered  as  equivalent  to  a  pressure  of  1  lb.  per  sq.  in. 

49.  Pressure  Due  to  Liquid  on  Submerged  Surface. — In  Fig.  7, 
let  abed  be  a  flat  plate  submerged  in  the  water  contained  in  the 
tank  T.  It  is  evident  that  the  pressure  (specific)  at  the  edge  ab  is 
less  than  it  is  at  the  edge  dc.  Let  /  be  the  center  of  gravity  of  the 
plate;  then  the  specific  pres- 
sure at  /  will  be  the  same  at 
any  point  along  a  horizontal 
line  drawn  on  the  plate  and 
passing  through  /,  because 
every  point  on  such  a  line 
will  be  at  the  same  depth 
below  the  level  of  the  liquid. 
Further,  this  specific  pressure 
will  be  the  average  pressure 
on  the  plate,  and  when  multiplied  by  the  area  of  the  plate,  the 
product  will  be  the  total  normal  pressure  P  on  the  plate.  Let 
h  —  the  depth  of  /  below  the  surface  of  the  liquid,  a  =  area  of 
plate,  and  P  =  the  total  normal  pressure  on  the  plate;  then, 

P  =  ahw  (1) 

in  which  w  is  the  specific  weight  (weight  of  a  unit  cube)  of  the 
liquid.  In  this  formula,  if  h  is  in  feet  and  a  is  in  square  feet,  w  is 
the  weight  of  a  cubic  foot;  hence,  for  water 

P  =  62Aah  (2) 

If  h  is  in  feet,  a  in  square  inches,  and  w  =  weight  of  1  cu.  ft. 

_  ahw  ,  . 

F  "  144  {3) 

and  for  water, 

P  =  Ay3ah  =  Mali  (4) 

For  a  curved  surface,  a  must  be  equal  to  the  projection  of  the 
surface  on  a  plane  perpendicular  to  the  normal  through  the 
center  of  gravity. 

Rule. — The  total  normal  pressure  upon  any  submerged  surface 
due  to  the  weight  of  the  liquid  is  equal  to  the  weight  of  a  prism  of  the 
liquid  whose  base  is  equal  to  the  projection  of  the  surface  on  a  plane 
perpendicular  to  the  normal  at  the  center  of  gravity  of  the  surface 
and  whose  altitude  is  the  depth  of  this  center  of  gravity  below  the 
surface  of  the  liquid. 


28 


ELEMENTS  OF  PHYSICS 


§4 


50.  From  the  foregoing,  it  will  be  evident  that  the  pressure  on 
the  bottom  of  a  vessel  has  nothing  to  do  with  the  shape  of  the 
vessel  that  contains  the  liquid.  Thus,  referring  to  Fig.  8,  if  the 
areas  of  the  bottoms  ab  of  the  four  vessels  shown  is  the  same,  and 
the  depth  of  the  water  is  the  same  in  all  the  vessels,  the  pressure 
(total  pressure)  on  the  bottoms  of  all  the  vessels  will  be  the  same, 
since  their  centers  of  gravity  are  at  the  same  depth  below  the 


T 


IT 


Fig.  9. 

surface  of  the  liquid.  Further,  if  all  the  vessels  stand  on  a  com- 
mon flat,  level  surface,  and  the  height  of  the  liquid  in  each  is 
the  same,  then  if  all  are  filled  with  the  same  liquid  and  are  con- 
nected together  by  a  pipe,  the  level  of  the  liquid  will  still  be  at  the 
same  distance  above  the  bases.  This  fact  is  strikingly  shown  in 
Fig.  9,  where  the  vessel  B  has  a  much  larger  cross-section  than  A . 
If  water,  for  example,  be  poured  into  A  (or  B),  it  will  flow  into 
B  {A)  through  the  connecting  pipe,  and  when  the  pouring  is 
stopped,  it  will  be  found  that  the  water  is  at  the  same  level  in 
both  vessels.     This  phenomenon  is  expressed  in  the  familiar 

da'  : 


■ 

m    ftp 

■ 

m     tiM, 

E 

-h:-- 

i 

<*) 

\ 

Fig 

10. 

saying  "Water  seeks  its  level."  If  the  water  level  were  not  the 
same  in  both  vessels,  there  would  be  a  greater  specific  pressure  on 
the  bottom  and  at  the  entrance  to  the  pipe  in  one  vessel  than  in 
the  other,  and  the  water  would  flow  from  the  place  of  higher  pres- 
sure than  to  that  of  the  lower. 

51.  It  is  to  be  again  emphasized  that  the  total  normal  pressure 
on  any  submerged  surface  does  not  depend  upon  the  shape  of  the 


HYDROSTATICS 


29 


t 


vessel  that  contains  the  liquid.  Referring  to  (a),  Fig.  10,  suppose 
debagef  to  be  a  dam,  the  top  and  bottom  having  the  shape  of  a 
rectangle.  Then  the  total  normal  pressure  on  the  inside  face  of 
the  dam  depends  only  on  the  projected  area  of  that  face  and  the 
depth  of  the  center  of  gravity  of  the  face  below  the  surface  of  the 
liquid,  which  is  assumed  to  be  level  with  the  top  of  the  dam;  and 
it  makes  no  difference  how  far  back  the  water  may  extend.  It 
also  makes  no  difference  what  shape  of  the  cross-sec- 
tion of  the  dam  may  be,  provided  the  projected 
area  is  the  same;  that  is,  the  dam  may  be 
straight,  as  in  (a),  or  curved,  as  in  (6). 

52.  Combined  Pressures. — If  the  upper  sur- 
face of  a  liquid  is  free,  the  pressure  at  any  point 
in  the  liquid  is  that  due  to  the  depth  of  the  water 
(liquid)  at  that  point;  but  if  the  liquid  is  sub- 
jected to  an  external  pressure,  the  total  specific 
pressure  at  any  point  is  equal  to  the  sum  of  the 
specific  pressure  due  to  the  depth  of  the  point 
below  the  surface  of  the  liquid  and  the  specific 
pressure  due  to  the  external  pressure. 

Example.— ^Referring    to    Fig.     11,    A    represents   a 
cylinder  filled  with  water,  the  inside  dimensions  being, 
say,  18  in.  diameter  and  30  inches  long.     A  pipe  B,  %  in. 
in  diameter  is  connected  to  the  cylinder  at  b  and  is  filled 
to  the  point  a  with  water.      If  a  pressure  of  16  pounds 
be  applied  to  the  handle  C,  thus  pushing  on  a 
piston  touching  the  water  in  the  pipe  at  a, 
what  will  be  the  total  pressure  on  the  bottom 
of  the  cylinder?  on  the  top  of  the  cylinder? 
the   lateral    specific   pressure   at    b?     Let  the 
vertical    distance    h  between  the   top   of   the 
water  in  the  tube  and  the  bottom  of  the  vessel 
be  8  ft.,  and  between  b  and  the  bottom  of  the 
vessel  12  inches. 

Solution. — The  specific  pressure  on  the 
bottom  of  the  vessel  due  to  the  water  in  the 
cylinder  and  pipe  is,  by  formula  (2),  Art.  48, 
since  8  ft.  =  96  in.,  p  =  #„  X  96  =  3.4667  lb.  per  sq.  in. 

The  specific  pressure  on  the  water  at  any  point  in  the  cylinder  due  to  the 

push  on  the  handle  is  equal  to  the  pressure  on  the  piston  divided  by  the 

,    ,  16 

area  of  the  piston,  or  -„.^        -2  =  81.487  lb.  per  sq.  in.     The  total  specific 

pressure  on  the  bottom  is  81.487  +  3.467  =  84.954  lb.    per  sq.  in.,  and 
total  pressure  on  bottom  is  .7854  X  182  X  84.954  =  21,618  lb.     Ans. 
The  specific  pressure  on  the  top  due  to  the  water  in  the  pipe  is  p  =  jVb 


Fig.  11. 


30  ELEMENTS  OF  PHYSICS  §4 

X  (96  —  30)  =  2.383  lb.  per  sq.  in.,  and  the  total  specific  pressure  on  the  top 
is  81.487  +  2.383  =  83.87  lb.  per  sq.  in.  The  total  pressure  on  the  top  is 
.7854  X  18*  X  83.87  =  21,342  lb.     Ans. 

The  lateral  specific  pressure  at  b  is  &r(96  -  12)  =  3.033  lb.  per  sq.  in. 
due  to  the  water  only.  The  total  specific  pressure  at  b  is  81.487  +  3.033 
=  84.52  lb.  per  sq.  in.     in-s. 

Observe  that  the  specific  pressure  due  to  the  push  on  the  handle  is 
transmitted  undiminished  in  all  directions  within  the  cylinder,  while  that 
due  only  to  the  water  depends  on  the  depth  of  the  point  considered  below 
the  level  a. 


EXAMPLES 

(1)  Referring  to  Fig.  2,  suppose  the  weight  on  top  of  the  piston  is  1200 
pounds,  the  diameter  of  the  piston  is  21  in.,  and  the  depth  of  the  water  under 
the  piston  is  40  in.,  what  is  {a)  the  total  pressure  on  the  bottom  of  the 
cylinder?  (b)  on  the  bottom  of  the  piston?  (c)  the  specific  pressure  due  to 
the  weight.  f  (o)  1700.3  1b. 

Ans.l  (c)  12001b. 

I  (c)  3.4646 -lb.  per  sq.  in. 

(2)  Referring  to  the  preceding  example,  what  is  the  force  (total  normal 
pressure)  tending  to  separate  one  half  of  the  cylinder  from  the  other  half? 

Ans.  3516.91b. 

(3)  In  Fig.  9,  suppose  that  the  diameter  of  A  is  1.72  in.  and  of  B  9.8  in. 
If  a  piston  weighing  12  lb.  rests  on  top  of  the  water  in  B  and  a  force  of 
24  lb.  (including  weight  of  small  piston)  is  applied  to  a  piston  on  top  of 
the  water  in  A,  what  force  (weight)  must  be  applied  to  the  piston  in  B  to 
keep  it  stationary?  Ans.  767.12  lb. 

Suggestion. — The  weight  of  the  piston  in  B  must  be  subtracted  from  the 
upward  pressure  of  the  water  to  find  the  total  downward  force  required  to 
balance  the  pressure  on  piston  in  A. 

(4)  Neglecting  the  weight  of  the  water  in  the  last  example,  how  far  will 
the  piston  in  B  move  when  the  piston  in  A  moves  3L4  in.? 

Ans.  0.1  in.  very  nearly. 

(5)  Referring  to  Fig.  7,  suppose  abed  to  be  a  flat  rectangular  plate  SL2  hi. 
by  11  in.,  and  that  its  center  of  gravity  is  43  in.  below  the  water  level;  what 
is  the  total  normal  pressure  on  the  plate?  Ans.  145.18  lb. 

(6)  Referring  to  the  example  of  Art.  52,  what  will  be  the  total  upward 
pressure  against  the  top  of  the  cylinder,  if  the  diameter  of  the  pipe  B  is  2  in., 
the  other  dimensions  and  the  pressure  on  the  handle  C  being  the  same  as 
before?  Ans.  329. S3 1  lb. 

(7)  The  total  difference  of  level  between  the  top  of  the  water  in  a  reservoir 
and  the  nozzle  of  a  fire  hose  is  225  ft.;  what  is  the  pressure  of  the  water  at 
the  nozzle  when  the  water  is  not  flowing?  Ans.  97.5  lb.  per  sq.  in. 

(8)  A  certain  reservoir  has  a  uniform  cross-section  shaped  like  the  second 
illustration  in  Fig.  8.  The  bottom  is  a  rectangle  144  ft.  by  350  ft.;  what  is 
the  total  pressure  on  the  bottom  when  the  depth  of  the  water  is  75  ft., 
assuming  that  the  bottom  is  level? 

Ans.  235,872,000  lb.  =  117,936  tons. 


§4  HYDROSTATICS 

BUOYANCY  AND  SPECIFIC  GRAVITY 


31 


BUOYANCY 

53.  Conditions  under  which  Bodies  Float  or  Sink. — All  bodies 
have  a  tendency  to  float;  let  us  consider  the  reason.  Referring 
to  Fig.  12,  ABCDEFG  represents  a  tank  filled  with  a  liquid  in 
which  two  bodies  M  and  N  are  immersed.  M  is  a  prism,  its 
bases  being  parallel  to  the  upper  (flat)  surface  of  the  liquid. 
The  downward  pressure  on  the  body  M  is  equal  to  the  weight  of  a 
prism  of  the  liquid  whose  volume  is  a'b'c'd'abcd;  the  upward 
pressure  is  the  weight  of  a  prism  of  the  liquid  whose  volume 
is  indicated  by  a'b'c'd'efg;  the  difference  of  these  volumes  is 
the  volume  of  the  prism;  and  the  difference  between  the  down- 
ward and  upward  pressures  is  equal  to  the  weight  of  a  prism  of 


Fig.  12. 

the  fluid  having  the  same  volume  as  the  prism  M .  In  other 
words,  the  difference  between  the  downward  and  upward  pressures 
is  equal  to  the  weight  of  the  liquid  displaced  by  the  body,  and  this 
statement  is  true  whatever  the  shape  of  the  body.  For  instance, 
let  N  be  an  irregular  body,  say  a  stone;  let  e'f'g'  be  the  projection 
of  the  stone  on  the  upper  surface  of  the  liquid;  then  the  down- 
ward pressure  is  equal  to  the  weight  of  a  column  of  the  liquid 
having  the  shape  e'f'g' edg;  the  upward  pressure  is  the  weight 
of  a  column  of  the  liquid  having  the  shape  e'f'g'efg;  and  the  dif- 
ference between  these  pressures  is  the  weight  of  a  body  of  fluid 
having  the  same  shape  as  the  stone.  Further,  it  makes  no  dif- 
ference how  far  below  the  surface  of  the  liquid  the  body  lies, 
the  weight  of  a  body  of  liquid  having  the  shape  of  the  submerged 
body  will  be  the  pressure  tending  to  force  the  body  upward. 


32  ELEMENTS  OF  PHYSICS  §4 

Since  gravity  tends  to  pull  the  body  down  and  the  difference 
between  the  downward  and  upward  pressures  tends  to  push  it 
up,  the  force  urging  the  body  down  through  the  liquid  is  equal  to 
the  weight  of  the  body  minus  the  weight  of  an  equal  volume  of 
the  liquid.  If,  therefore,  the  density  of  the  body  is  the  same  as 
that  of  the  liquid,  the  body  will  stay  in  any  position  in  the  liquid 
and  at  any  depth  that  it  may  be  placed,  assuming  the  density  of 
the  liquid  to  be  uniform.  If  the  density  of  the  body  is  less  than 
that  of  the  liquid,  the  body  will  float,  that  is,  only  a  part  of  it 
will  be  submerged,  the  weight  of  a  volume  of  liquid  equal  to  the 
submerged  part  being  equal  to  the  weight  of  the  body.  If  the 
density  of  the  body  is  greater  than  that  of  the  submerged  part, 
the  bod}^  will  sink,  that  is,  it  will  fall  through  the  liquid  until  it 
touches  bottom. 

As  an  illustration  of  the  statements  in  the  last  paragraph,  place 
an  egg  in  a  can  of  water;  the  egg  will  sink,  showing  that  its 
density  is  greater  than  water.  Now,  leaving  the  egg  in  the  water, 
dissolve  in  it  some  salt  (or  sugar),  stirring  the  water  so  it  will 
be  equally  dense  throughout.  The  salt  being  denser  than  the 
water,  the  density  of  the  water  will  gradually  increase  until  it 
becomes  the  same  as  that  of  the  egg,  and  when  this  point  is 
reached,  the  egg  will  stay  in  any  position  in  the  water  and  at 
any  depth.  As  more  salt  is  added  and  dissolved,  the  water 
becomes  denser  than  the  egg,  and  the  egg  will  rise  and  float,  a 
part  of  it  extending  out  of  the  water. 

54.  The  tendency  of  any  body  to  float  when  immersed  in  a 
fluid  is  called  buoyancy,  and  the  denser  the  body  the  less  buoyant 
it  is.  What  is  true  in  this  respect  of  a  liquid  is  equally  true  of  a 
gas,  the  upward  or  buoyant  force  being  equal  to  the  weight  of  a 
body  of  gas  having  the  same  volume  as  the  submerged  body. 
In  the  case  of  a  balloon,  the  gas  (hydrogen)  with  which  it  is  filled 
is  very  much  lighter  than  air,  a  cubic  foot  of  air  weighing  about 
14.5  times  as  much  as  hydrogen.  Assuming  that  a  cubic  foot 
of  air  weighs  .08  pound  under  certain  conditions,  100,000  cubic 
feet  will  weigh  8000  pounds,  and  100,000  cubic  feet  of  hydrogen 
under  the  same  conditions  will  weigh  8000  4-  14.5  =  552  —  pounds. 
The  difference  is  8000  —  552  =  7448  pounds,  which  is  the  buoyant 
effect,  or  force  urging  the  balloon  upward.  The  total  upward  force 
is  8000  pounds,  but  this  is  counteracted  by  the  weight  of  the 
hydrogen,  the  weight  of  the  balloon,  and  the  load  lifted.  As  the 
balloon  moves  up,  the  air  becomes  less  and  less  dense,  and  a  point 


§4  HYDROSTATICS  33 

will  be  reached  where  the  balloon  can  ascend  no  higher,  remaining 
stationary  (in  still  air)  unless  some  of  the  hydrogen  be  allowed  to 
escape,  when  the  balloon  will  descend;  or,  if  the  balloon  carries 
ballast  and  this  be  thrown  out,  the  balloon  will  rise  higher. 

Mercury,  which  is  a  liquid  at  ordinary  temperatures,  is  so 
dense  that  iron,  copper,  lead,  etc.  will  float  in  it;  but  gold, 
platinum,  tungsten,  etc.  will  sink  in  it,  because  their  densities 
are  greater  than  the  density  of  mercury. 

55.  Weight  in  a  Vacuum. — A  vacuum  is  a  closed  space  from 
which  all  air  or  other  gas  has  been  removed;  the  inside  of  a 
carbon  filament  lamp  is  a  good  example  of  a  vacuum.  The  air 
must  be  removed,  since  it  would  otherwise  unite  with  the  carbon 
(when  heated)  and  the  filament  would  be  destroyed.  A  little 
consideration  will  show  that  a  body  will  weigh  more  in  a  vacuum 
than  in  air,  because  the  buoyant  effect  of  the  air  counteracts 
by  just  that  much  the  force  of  gravity.  For  example,  if  a  cubic 
foot  of  water  weighs  62.4  pounds  in  air  when  the  air  weighs 
.08  lb.  per  cu.  ft.,  a  cubic  foot  of  water  will  weigh  62.4  +  -08 
=  62.48  pounds  in  a  vacuum.  The  weight  in  a  vacuum  is  there- 
fore the  true  weight  of  a  body.  However,  in  all  practical  and 
commercial  transactions,  the  weight  of  a  body  in  air  is  the  weight 
that  is  used;  it  is  only  in  accurate  scientific  calculations  that 
weights  are  expressed  as  the  true  weights  in  a  vacuum. 


SPECIFIC  GRAVITY 

56.  Definition. — The  specific  gravity  of  a  body  is  the  ratio  of 
the  weight  of  the  body  to  the  weight  of  an  equal  volume  of  water. 
Thus,  the  weight  of  a  cubic  foot  of  cast  iron  is  commonly  taken 
as  450  pounds,  and  (when  finding  the  specific  gravity)  the  weight 
of  a  cubic  foot  of  water  is  62.4  pounds;  hence,  the  specific 

450 
gravity  of  cast  iron  is  t^w;  =  7.21.     In  other  words,  cast  iron 

weighs  7.21  times  as  much  as  an  equal  volume  of  water  at  its 
temperature  of  maximum  density.  When  62.4  pounds  is  used 
as  the  weight  of  a  cubic  foot  of  water  in  finding  the  specific 
gravity,  it  is  useless  to  find  the  value  of  the  ratio  to  more  than 
three  significant  figures. 

62.4 
The  specific  gravity  of  water  is  evidently  1,  since  ~^  =  1. 

Consequently,  if  the  specific  gravity  of  a  body  is  greater  than  1, 


34  ELEMENTS  OF  PHYSICS  §4 

the  density  of  the  body  is  greater  than  the  density  of  water,  and 
the  body  will  sink  in  water;  but  if  the  specific  gravity  is  less  than 
1,  the  density  of  the  body  is  less  than  that  of  water,  and  the  body 
will  float  in  water. 

57.  Let  s  =  the  specific  gravity  of  a  body,  W  =  its  weight, 
and  w  =  weight  of  an  equal  volume  of  water;  then, 

w  m 

s  =  —  (1) 

w 
W  =  wa  (2) 

The  specific  gravity  of  most  materials  and  substances  can  be 
obtained  from  printed  tables;  hence,  if  the  volume  and  specific 
gravity  of  any  body  or  substance  is  known,  its  weight  can  read- 
ily be  found.  For,  the  weight  of  an  equal  volume  of  water  is 
62.4  V,  where  V  is  the  volume  in  cubic  feet,  and  this  equals  w 
in  formula  (2);  therefore,  substituting  in  formula  (2), 

W  =  62.4  Vs  (3) 

If  v  =  the  volume  in  cubic  inches,  1  cu.  in.  of  water  weighs 
s\%    pounds    (see    Art.    48),    and 

W  =  —  (4) 

360  w 

Example. — Taking  the  specific  gravity  of  silver  as  10.53,  what  is  the 

weight  of  8.65  cu.  in.? 

Solution. — Applying  formula  (4),  v  =  8.65  and  s  =  10.53;  hence, 

w       13  X  8.65  X  10.53  ocn 

W  =  sttx =  3.289  lb.     Am. 

dot) 

68.  When  using  the  metric  system  of  weights  and  measures 

in  connection  with  specific  gravity  problems,  the  cubic  decimeter 

is  taken  as  the  standard  of  volume.     Since  1  cubic  decimeter 

=  1  liter,  and  since,  by  definition,  1  kilogram  is  the  weight  of  1 

liter  of  water,  then  letting  W  —  the  weight  of  one  cubic  decimeter 

W 

of  the  substance,  formula  (1)  of  the  last  article  becomes  8  =  -y 

=  W;  that  is,  the  specific  gravity  of  any  substance  is  numerically 

equal  to  the  weight  in  kilograms  of  1  cubic  decimeter  of  the 

substance.     As  an  illustration,  the  weight  of  1   cubic  foot  of 

cast  iron  is  450  pounds;  since  1  kg.  =  2.2046  pounds  and  1  cu. 

a  ftino^         -i         a        t       +  •  •  i,      450X61.024 

dm.  =  61.024  cu.  in.,  1  cu.  dm.  of  cast  iron  weighs  ,^0  w  n  m4„ 

a      1728  X  2.2046 

=  7.21+  kg.     The    value    previously    found    for    the    specific 

gravity  of  cast  iron  was  7.21;  hence,  the  numerical  values  7.21 

agree. 


§4  HYDROSTATICS  35 

To  find  the  specific  gravity  of  a  liquid,  all  that  is  necessary  is 
to  fill  a  liter  measure,  weigh  it,  and  then  subtract  the  weight  of 
the  measuring  vessel;  the  result  expressed  in  kilograms  will  be 
the    numerical    value    of    the    specific    gravity    of    the  liquid. 

To  find  the  specific  gravity  of  a  solid,  it  would  be  quite  dif- 
ficult to  form  it  into  the  shape  of  a  cube  measuring  1  cu.  ft.  or  1 
cu.  dm.  on  each  edge;  in  fact,  in  many  cases  it  would  be  impossible. 
Consequently,  the  usual  method  is  to  immerse  the  object  in 
water  and  note  the  loss  in  weight.  This  loss  in  weight  is  evidently 
equal  to  the  weight  of  a  volume  of  water  equal  to  the  volume  of 
the  solid,  the  value  of  W  in  formula  (1),  Art.  57.  Let  W 
=  the  weight  in  air  and  W  =  the    weight    in    water;    then 

W 

s  -  w  -  w 

For  example,  suppose  a  certain  small  object  weighs  367  grains 
in   air   and   310  grains  in   water;   its  specific  gravity  is  s  = 

367  -  310       °  " 

59.  Distinction    between    Density    and    Specific    Gravity. — 

Density  may  be  defined  as  specific  mass  or  as  specific  weight, 
according  to  which  of  the  two  definitions  of  Art.  30  is  used. 
The  word  specific,  as  used  in  physics,  always  implies  a  reference 
to  some  standard.  Thus,  specific  weight  means  the  weight  of  a 
unit  of  volume  (1  cu.  ft.,  1  cu.  dm.,  etc.),  specific  pressure  is  the 
pressure  per  unit  of  area  (1  sq.  ft.,  1  sq.  cm.,  etc.),  specific  volume 
is  the  volume  of  a  unit  of  weight  (1  lb.,  1  kg.,  etc.).  If,  therefore, 
density  be  defined  as  the  weight  of  a  unit  of  volume  and  the  unit 
of  volume  be  taken  as  1  cu.  dm.  =  1  liter,  the  numerical  values 
of  the  density  and  specific  gravity  of  any  body  will  be  equal;  in 
fact,  many  writers  on  scientific  subjects  use  these  two  terms 
interchangeably.  They  do  not,  however,  mean  exactly  the  same 
thing,  since  density  means  specific  weight  (or  specific  mass), 
while  specific  gravity  means  the  number  of  times  heavier  a 
substance  is  than  an  equal  volume  of  water.  To  prevent  any 
ambiguity,  the  first  definition  of  density,  according  to  which 

D  =  y  =  -y>  will  be  used  hereafter,  except  for  gases. 

60.  Specific  Gravity  of  Gases. — Gases  are  so  much  lighter 
than  equal  volumes  of  liquids  and  solids  that  is  customary  to 
express  their  specific  gravities  as  the  ratio  of  the  weight  of  any 
volume  of  gas  to  the  weight  of  an  equal  volume  of  air.     The 


36 


ELEMENTS  OF  PHYSICS 


§4 


weight  of  any  gas  depends  not  only  on  its  volume  but  also  on  its 
temperature  and  pressure.  The  weight  of  a  cubic  foot  of  air  at 
32°  F.  (0°  C.)  when  its  pressure  is  14.695  pounds  per  sq.  in.  is 
.08071  pound.  (This  combination  of  pressure  and  temperature 
is  called  standard  conditions.)  Hence,  if  the  specific  gravity 
of  nitrogen  be  given  in  a  certain  table  as  .970,  the  weight  of  a  cubic 
foot  at  32°  F.  and  a  pressure  of  14.695  lb.  per  sq.  in.  is  .08071 
X  .970  =  .07829  pound.  Properties  of  gases  will  be  considered 
later. 


HYDROMETERS 

61.  Definition. — A  hydrometer  is  an  instrument  by  means  of 
which  the  specific  gravity  of  a  liquid  may 
be  found;  they  are  made  in  many  forms 
and  are  called  by  various  names,  but  they 
all  depend  for  their  action  on  the  fact  that 
if  a  body  lighter  than  the  liquid  in  which  it 
is  placed  sink  to  a  certain  mark  on  the 
body,  it  will  sink  farther  in  a  lighter  liquid 
(one  less  dense)  and  not  so  far  in  a  heavier 
liquid  (one  of  greater  density).  Fig.  13 
shows  two  forms  of  the  instrument,  the 
one  at  (a)  being  for  liquids  lighter  than 
water  and  the  one  at  (b)  for  liquids  heavier 
than  water.  As  will  be  seen,  they  consist 
of  glass  tubes  closed  at  both  ends  and  con- 
taining a  graduated  scale.  One  end  is 
loaded  with  mercury  or  shot  to  make  the 
instrument  stand  upright  when  placed  in 
the  liquid.  The  enlarged  part  above  the 
loaded  end  increases  the  buoyancy.  The 
instrument  shown  at  (a)  has  a  narrow  stem, 
increasing  its  sensitiveness,  and  adapting 
it  to  liquids  heavier  than  water;  the  one 
shown  at  (6)  has  a  wide  stem,  and  is  for 
use  in  liquids  lighter  than  water. 

The  density  of  water  and  other  liquids 
usually  decreases  as  the  temperature  in- 
Pjq  13  creases;  for  this  reason  it  is  necessary  to 

have  a  standard  temperature  for  the  gradu- 
ation of  the  hydrometer  and  for  the  liquid  when  the  hydro- 


§4  HYDROSTATICS  37 

meter  is  placed  in  it:  this  temperature  is  generally  15° C.  or  60° F. 
C.  and  F.  mean  centigrade  and  Fahrenheit  respectively,  and  will 
be  explained  later.  Although  15°  C.  really  equals  59°  F.,  the 
difference  between  59°  and  60°  is  so  slight  that  it  may  usually 
be  neglected. 

62.  The  Graduations  on  the  Scales. — The  main  graduations 
on  any  hydrometer  scale  are  called  degrees,  and  these  are  fre- 
quently subdivided  into  tenths  of  a  degree.  Some  instruments  are 
so  divided  that  they  give  the  specific  gravity  while  in  others,  if 
the  specific  gravity  be  desired,  it  must  be  calculated  from  the  scale 
,  reading.  The  scale  most  commonly  used  in  the  United  States  and 
Canada  is  the  Beaume;  and  while  there  are  several  formulas  for 
calculating  the  Beaume  scale,  the  so-called  American  Beaume 
is  the  only  one  that  will  be  considered  here.  Any  reference  to  the 
Beaume  scale  in  this  course  will  be  understood  to  be  the  American 
Beaume  at  60°  F. 

Let  s  =  the  specific  gravity  of  the  liquid  to  be  tested  at  60°  F., 
and  let  5  =  the  reading  of  the  scale  in  degrees  Beaume;  then 
for  liquids  lighter  than  water, 

140  m 

S  ~  130  +  B  {L) 

For  liquids  heavier  than  water, 

S       145  -B  {  } 

For  liquids  lighter  than  water,  the  scale  evidently  begins  at 

140 
10,    since    by    formula    (1)    s  =     ^  =1    =  the    specific 

gravity  of  water.     And  for  liquids  heavier  than  water,  the  scale 

145 
begins    at    0,    since    by   formula    (2),    s  =     ,„  _  »  =  1. 

To  use  the  instrument,  partly  fill  a  glass  container,  of  such 
depth  that  the  hydrometer  will  not  touch  bottom,  with  the  liquid 
to  be  tested,  first  being  sure  that  its  temperature  is  60°,  then 
gently  insert  the  hydrometer,  and  when  it  conies  to  rest  in  an 
upright  position,  note  where  the  upper  surface  of  the  liquid 
crosses  the  scale;  this  will  be  the  reading  in  degrees  Beaume.  If 
the  specific  gravity  is  desired  (which  is  not  usually  the  case,  the 
reading  in  degrees  Beaume  being  generally  sufficient),  calculate 
it  by  substituting  the  reading  for  B  in  formula  (1)  or  (2). 

If  it  be  desired  to  convert  specific  gravity  into  degrees  Beaum6, 


38  ELEMENTS  OF  PHYSICS  §4 

simply  solve  the  above  formulas  for  B,  obtaining  for  liquids 
lighter    than    water, 

B  =  ™-=J*9*  (3) 

and  for  liquids  heavier  than  water, 

=  145(«  -  1)  ( 

S 

Example  1. — Suppose  a  certain  liquid  that  is  known  to  be  heavier  than 
water  has  a  density  of  18.4°  Beaume;  what  is  its  specific  gravity? 
Solution. — Substituting  18.4  for  B  in  formula  (2), 

s  —  tt; ,0  .    =  1.1453.     Ans. 

145  —  18.4 

Example  2. — Knowing  that  the  specific  gravity  of  a  certain  liquid  is  .886, 

what  is  the  density  in  degrees  Beaume? 

Solution. — Since  the  specific  gravity  is  less  than  1,  use  formula  (3),  and 

„       140  -  130  X  .8S6      __„„  .       . 

"  =  ooa. =  28  Beaum6.    Ans. 

•  oou 

63.  Object  of  Hydrometer. — The  reason  for  using  the  hydrom- 
eter to  get  the  comparative  densities  of  liquids  is  that  materials 
bought  and  sold  in  liquid  form  (mixtures  of  several  different 
materials)  do  not  contain  a  constant  percentage  of  some  particu- 
lar substance  for  which  the  liquid  was  bought.  As  a  simple 
illustration,  consider  a  mixture  of  common  salt  and  water;  the 
greater  the  percentage  of  salt  in  the  solution  the  greater  the  den- 
sity of  the  solution.  In  a  solution  of  salt  and  water,  if  the  den- 
sity of  the  solution  or  its  specific  gravity  be  known,  then  by  means 
of  a  special  formula  or  table,  the  percentage  of  salt  can  be  found. 
This  same  method  can  be  used  to  find  the  percentage  of  other 
substances  in  solution.  In  many  pulp  and  paper  mills,  the 
hydrometer  is  used  for  taking  the  strength  of  the  bleach  liquor; 
if  the  bleach  is  of  poor  quality,  that  is  if  it  contains  soluble  im- 
purities, as  calcium  chloride,  which  remains  over  when  the  bleach 
powder  decomposes,  misleading  results  will  be  obtained  with  the 
hydrometer,  because  the  calcium  chloride  increases  the  density 
just  as  the  bleach  itself  does.  Consequently,  before  dependence 
can  be  placed  on  the  results  obtained  with  the  aid  of  the  hy- 
drometer, one  must  be  sure  that  the  solution  itself  is  pure. 


EXAMPLES 

(1)  A  liter  of  hydrogen  under  standard  conditions  weighs  .0S9S73  gram, 
a  liter  of  air  weighs  1.2929  gram ;  (a)  what  is  the  specific  gravity  of  hydrogen  ? 
how  many  liters  of  hydrogen  equal  the  weight  of  1  liter  of  air? 

/  (a)    .069514 
AnS-\(b)   14.389 


§4 


HYDROSTATICS 


39 


(2)  A  flask  holding  50  c.c.  (cubic  centimeters)  is  filled  with  a  liquid  and 
weighed;  after  deducting  the  weight  of  the  flask,  the  weight  of  the  liquid  is 
found  to  be  64.6  g.  (grams);  (a)  what  is  the  specific  gravity  of  the  liquid? 
(6)  what  is  its  relative  density  in  degrees  Beaumc? 

,  ((a)   1.292 

AnS-    I  (6)  32.77°   Be. 

(3)  What  is  the  weight  of  a  cold  drawn  steel  rod  2%  in.  in  diameter  and 
13  ft.  long,  the  specific  gravity  of  the  steel  being  7.83?  Ans.  175.4  lb. 

(4)  The  specific  gravity  of  oxygen  is  1.1053;  what  is  the  weight  of  468 
cubic  feet  under  standard  conditions?  Ans.  37.77  lb. 

(5)  Suppose  a  liquid  that  is  lighter  than  water  has  a  relative  density  of 
42°  Beaume;  what  is  its  specific  gravity?  Ans.  .814. 

(6)  The  relative  density  of  a  certain  liquid  that  is  lighter  than  water  is 
17.4°Beaume;  what  is  the  weight  of  50  c.c?  Ans.  47.5  g. 


eh 


# 


CAPILLARITY 

64.  Capillary  Attraction. — The  word  capillary  means  hair-like, 
and  as  used  in  physics,  it  refers  to  small  or  fine  tube-like  holes  and 
channels.  If  a  clean  glass  rod  be  placed  in  a  vessel  of  water,  as 
shown  in  (a),  Fig.  14,  it  will  be  found  that  the  surface  of  the 
water  around  the  rod  is  not  level,  but  curved,  the  water  being 
drawn  up  about  the  rod,  and  the 
part  touching  the  rod  will  be  higher 
than  the  water  level  as  at  a  and  b. 
If,  instead  of  a  glass  rod,  a  glass 
tube  be  used,  the  water  in  the  tube 
will  be  higher  than  the  water  level 
outside  (as  at  c  in  Fig.  16);  the 
shape  of  the  upper  surface  of  the 
water  in  the  tube  will  also  be 
curved,  the  outer  edge  being  higher 
than  the  center.  It  will  also  be  found  that  the  smaller  the 
diameter  of  the  hole  the  higher  the  water  will  rise  in  the  tube. 
This  phenomenon  is  called  capillary  attraction,  and  the  results 
noted  are  caused  by  the  adhesion  of  the  water  to  the  tube  and 
the  cohesion  of  the  water  particles.  Fibers  used  in  making  paper 
are  really  fine  tubes  and  consequently  behave  similarly. 

If  the  glass  tube  be  covered  inside  and  out  with  a  thin  coating 
of  grease,  an  exactly  opposite  effect  will  be  obtained;  the  water 
will  be  depressed  inside  and  outside  of  the  tube,  as  shown  at  (6), 
Fig.  14,  and  the  curves  will  be  reversed.  This  latter  effect  will 
also  be  observed  if  a  clean  glass  tube  be  inserted  in  mercury. 


fft-yg 

Ik,-- 

HSl 

-_-  -  --_---_"_-_-_---:) 

(a) 


(b) 


Fig.  14. 


40 


ELEMENTS  OF  PHYSICS 


It  will  be  found  on  examination,  that  in  the  first  case,  the  liquid 
wet  the  tube  and  in  the  second  case,  it  did  not  wet  the  tube.  In 
general,  capillary  attraction  draws  the  liquid  up  when  the  liquid 
wets  the  surface  and  pushes  it  down  when  the  liquid  does  not  wet 
the  surface.  A  striking  illustration  may  be  obtained  by  means  of 
two  flat  panes  of  glass,  partly  immersed  in  a  vessel  of  water,  as 
shown  in  Fig.  15.  Bringing  two  of  the  ends  together  and  sep- 
arating the  other  two  ends  slightly,  as  shown,  the  water  will  be 
found  to  rise  in  a  curve,  the  highest  point  being  at  the  ends  that 
touch. 


Fig.  15. 


If  one  end  of  a  lump  of  sugar  be  touched  to  and  held  in  contact 
with  the  surface  of  a  liquid,  as  water,  milk,  etc.,  the  entire  lump 
will  soon  become  wet,  the  liquid  being  raised  through  the  pores  by 
capillary  attraction.  When  a  candle  burns,  the  wax  or  tallow  is 
melted  (changed  to  a  liquid)  and  capillary  attraction  draws  the 
melted  wax  or  tallow  up  through  the  wick  to  the  flame.  If  there 
were  no  wick,  the  candle  would  not  burn  in  the  regular  manner. 

65.  The  vertical  height  through  which  a  liquid  will  be  raised  by 
capillary  attraction  above  the  true  level  of  the  liquid  varies  in- 
versely-as  the  diameter  of  the  tube,  for  the  same  liquid.  Thus, 
water  will  rise  higher  than  alcohol,  but  for  the  same  liquid  in  two 
tubes  A  and  B,  if  A  is  twice  as  large  as  B,  the  liquid  will  rise  only 
one-half  as  far  in  A  as  in  B;  if  the  diameter  of  A  is  .03  in.  and  of  B 
.01  in.,  the  liquid  will  rise  .03  -*-  .01  =  3  times  as  far  in  B  as  in  A. 


§4 


HYDROSTATICS 


41 


Capillary  attraction  need  be  considered  only  when  the  diameter 
of  the  tube  is  quite  small.  The  curve  formed  by  the  top  of  the 
liquid  is  called  the  meniscus,  and  it  is  crescent  or  bow-shaped. 
The  meniscus  at  c,  Fig.  16,  is  said  to  be  concave  upward,  and  that 
in  (b),  Fig.  14,  is  said  to  be  convex  upward.  The  top  of  the 
meniscus,  in  either  case,  is  the  point  where  the  axis  of  the  tube 
intersects  it.  When  reading  the  height  of  a  column  of  fluid  that 
has  a  meniscus,  it  is  usual  to  take  the  top  of  the  meniscus  as  the 
top  of  the  column.  This  applies  to  thermometers,  barometers, 
pipettes,  burettes,  etc. 

An  interesting  example  of  capillary  attraction  is  shown  in  Fig. 
16.  Here  A  and  B  are  glass  vessels  connected  at  D.  C  is  a  small 
glass  tube  also  connected  to  D. 
The  diameters  of  A  and  B  are  such 
that  the  water  level  in  each  is  the 
same,  and  is  indicated  by  the  line 
ab.  The  water  level  in  C,  how- 
ever, is  much  higher,  being  indi- 
cated by  c.  The  distance  cd 
represents  the  vertical  height  that 
the  water  was  raised  by  capillary 
attraction,  and  it  does  not  in  any 
way  increase  the  hydrostatic  pres- 
sure on  the  water  in  A  and  B,  and  it  would  evidently  not  be 
correct  to  measure  the  height  of  c  in  any  calculation  regarding 
the  normal  pressure  at  any  point  of  the  liquid  in  the  vessels. 

66.  Some  Examples  of  Capillary  Action. — Capillary  action 
plays  a  very  important  part  in  ordinary  everyday  affairs.  With- 
out capillary  action,  no  plant  could  live,  since  capillary  attraction 
is  what  draws  the  sap  up  from  the  ground.  1^  is  capillary  attrac- 
tion that  enables  a  piece  of  blotting  paper  to  absorb  ink,  that 
brings  oil  to  the  flame  through  the  wick  of  a  lamp,  that  causes 
wood  to  swell  when  placed  in  water.  In  coloring  or  sizing  paper, 
capillary  attraction  helps  by  drawing  solutions  into  the  fibers. 
It  can  be  made  to  exert  an  immense  force.  Thus,  if  a  hemp  rope 
be  drawn  tight  and  is  then  wet  with  water,  it  contracts,  because 
the  fibers  run  around  in  helixes  (somewhat  like  a  screw  thread) ; 
as  the  capillary  attraction  draws  up  the  water,  the  rope  swells  and 
the  fibers  shorten  in  a  direction  lengthwise  of  the  rope  the  re- 
sult being  a  tremendous  pull. 


I 

lh 

1 

r~  ~"A 

c->5^-^^^»^-^v>-: 

Fig.  16. 


42  ELEMENTS  OF  PHYSICS  §4 


PNEUMATICS 


THERMOMETERS 

67.  Pneumatics  is  that  branch  of  science  that  treats  of  the 
properties  of  air  and  gases;  it  might  be  called  the  hjrdrostatics  of 
gases. 

As  was  previous^  pointed  out,  all  gases  completely  fill  the 
vessels  that  contain  them;  it  is  also  to  be  noted,  that  the  density 
of  a  gas  is  uniform  through  its  entire  extent,  that  is,  no  matter 
what  the  shape  of  the  container,  a  cubic  inch,  say,  of  the  gas  will 
have  the  same  density  regardless  of  what  part  of  the  vessel  it  is 
taken  from.  Theoretically,  this  last  statement  is  not  quite  true, 
since  a  cubic  inch  taken  from  the  top  of  a  vessel  will  weigh  a 
trifle  less  than  that  taken  from  the  bottom,  the  column  of  gas 
having  weight  in  the  same  manner  that  a  column  of  water  has 
weight;  but,  in  practice,  this  may  be  neglected,  and  the  density  of 
a  gas  may  be  considered  to  be  uniform  throughout  the  container. 

68.  Tension. — The  molecules  of  gases  are  in  rapid  vibration, 
and  they  are  constantly  trying  to  escape  from  the  vessel  that  con- 
tains the  gas;  the  result  is  that  the  gas  exerts  a  pressure  against 
the  walls  of  the  container  in  much  the  same  manner  that  a 
helical  spring  exerts  a  pressure  when  a  load  is  placed  on  it.  The 
condition  of  the  gas  or  spring  that  causes  it  to  exert  pressure  is 
called  tension.  The  force  with  which  the  spring  presses  against 
the  load  is  equal  to  and  measures  its  tension,  and  the  pressure 
which  gases  exert  against  the  walls  of  their  containers  also  equals 
and  measures  their  tension.  Tension  is  measured  in  the  same 
manner  and  in  the  same  units  as  pressure,  and  the  word  pressure 
is  frequently  used  instead  of  tension,  since  the  tension  must 
always  equal  the  pressure. 

69.  Before  the  tension  of  any  gas  can  be  found,  it  is  necessary, 
as  will  be  more  fully  explained  later,  to  know  its  temperature, 
because  any  increase  or  decrease  in  the  temperature  increases 
or  decreases  the  tension,  provided  the  volume  of  the  gas  does  not 
change.  Under  ordinary  conditions,  temperatures  of  bodies 
are  measured  by  instruments  called  thermometers. 

70.  Thermometer  Scale. — All  thermometers  depend  for  their 
operation  upon  the  fact  that  most  substances  expand  when  heated 


§4  PNEUMATICS  43 

and  contract  when  cooled;  if  this  variation  in  length,  area,  or 
volume  can  be  measured,  a  thermometer  can  be  constructed. 
The  substance  most  generally  used  for  this  purpose  is  mercury. 
As  usually  constructed,  a  glass  tube  with  a  bulb  at  one  end  is 
partly  filled  with  mercury;  the  air  is  then  exhausted  above  the 
mercury  and  the  other  end  of  the  tube  is  sealed.  When  the  tube 
is  heated,  the  mercury  expands,  and  the  column  in  the  tube 
lengthens;  and  when  the  tube  is  cooled,  the  mercury  contracts, 
and  the  column  shortens.  The  hole  in  the  tube  is  quite  fine,  so 
that  a  small  expansion  in  the  volume  of  the  bulb  will  make  a 
considerable  increase  in  the  length  of  the  column. 

In  constructing  a  scale  for  a  thermometer,  it  is  assumed  that  the 
ice  made  from  pure  water  always  melts  at  the  same  temperature 
and  that  pure  water  always  boils  at  the  same  temperature, 
when  the  pressure  conditions  are  the  same.  The  tube  is  therefore 
inserted  in  melting  ice,  and  the  height  of  the  mercury  column 
is  marked  on  the  scale.  It  is  next  inserted  in  boiling  water, 
and  the  height  of  the  mercury  column  is  marked  on  the  scale. 
It  is  now  assumed  that  if  the  distance  between  these  two  points 
be  divided  into  equal  parts,  each  part  will  indicate  an  equal  rise 
(or  fall)  in  temperature.  Experience  has  shown  these  assumptions 
to  be  correct. 

There  are  two  thermometer  scales  in  general  use — the  Fahren- 
heit scale  and  the  centigrade  or  Celsius  scale.  The  former  is 
in  common  use  in  English  speaking  countries,  and  the  latter  in 
all  other  countries  (except  Russia)  and  by  scientists  generally. 
In  both  scales,  the  principle  divisions  are  called  degrees,  and 
these  are  subdivided  into  tenths,  hundredths,  etc.,  the  degrees 
on  the  Fahrenheit  scale  being  denoted  by  N°F.  and  those  on 
the  centigrade  scale  by  N°C,  N  representing  the  number  of 
degrees;  thus,  59°F.  and  15°C,  represent  59  degrees  Fahrenheit 
and  15  degrees  centigrade,  respectively. 

71.  When  graduating  a  thermometer  (thermometer  means 
heat  measurer)  in  accordance  with  the  Fahrenheit  scale,  the 
point  that  indicates  the  temperature  of  melting  ice  is  marked 
32°,  and  the  point  that  indicates  the  temperature  of  boiling 
water  is  marked  212°;  the  difference  is  212°  -  32°  =  180°, 
and  the  distance  between  these  two  points  is  divided  into  180 
equal  parts,  each  of  which  is  1°F.  These  divisions  are  carried 
above  and  below  the  two  fixed  points,  those  below  being  numbered 
29,  28,  etc.  until  0  is  reached,  which  will  be  32°  below  the  melting 


44  ELEMENTS  OF  PHYSICS  §4 

point  of  ice.  All  graduations  below  0  are  negative,  and  they 
increase  numerically  from  1  up;  thus  —  12°F.  means  12  degrees 
below  zero  on  the  Fahrenheit  scale. 

The  centigrade  scale  (sometimes  called  the  Celsius  scale)  is 
graduated  according  to  the  same  plan.  The  point  indicating  the 
temperature  of  melting  ice  is  marked  0°;  the  point  indicating 
the  temperature  of  boiling  water  is  marked  100°;  and  the  distance 
between  them  is  divided  into  100  equal  parts.  Since  this  distance 
is  the  same  on  both  scales,  1°C:  1°F.  =  180  :  100  from  which 
it   is   seen  that    1°C.  =  \U  =  |F°;  and  1°F.  =  4Jjjj  =  |-C°. 

72.  To  convert  degrees  C.  into  degrees  F.,  multiply  the  reading 

in  degrees  C.  by  f,   and  the  result  will  be  the  number  of  F° 

above  the  temperature  of  melting  ice;  to  find  the  temperature 

above  0°F.,  add  32°  to  the  product.     Expressed  as  a  formula 

F.°  =  C.°X   I  4-  32  (1) 

To  convert  degrees  F.  into  degrees  C,  subtract  32°  from  the 
reading  and  multiply  the  remainder  by  -§;  the  product  will  be  the 
degrees  C.     Expressed  as  a  formula. 

C°.  =  (F°.  -  32°)  Xf  (2) 

Example  1. — How  many  degrees  F.  are  equivalent  (a)  to  15°C?  (b) 
to  -20°C? 

Solution.— (a)  Applying  formula  (1),  15  X  I  +  32  =  59°F.     Ans. 

(b)  Applying  formula  (1),  -20  X  g  +  32  =  -3G  +  32  =  -4°F.     Ans. 

Example  2. — (a)  How  many  degrees  C.  are  equal  to  240°F.?  (6)  to 
—  22°F? 

Solution.— (a)  Applying  formula  (2),  (240  -  32)  X  f  =  115E°C.  Ans. 
(6)  Applying  formula  (2),  (-22  -  32)  X  I  =  -  30°C.     Ans. 

In  both  of  the  above  examples,  the  numbers  given  are  supposed  to  be 
thermometer  readings. 


THE  ATMOSPHERE 

73.  Perfect  and  Imperfect  Gases. — A  perfect  gas  is  one  that 
remains  a  gas  under  all  conditions  of  temperature  and  pressure; 
such  a  gas  is  also  called  a  permanent  gas.  There  is  really  no 
such  thing  as  a  perfect  gas,  since  all  gases  have  been  liquefied 
at  extremely  low  temperatures  and  under  very  high  pressures. 
For  practical  purposes,  however,  hydrogen,  nitrogen,  oxygen, 
atmospheric  air,  etc.  may  be  considered  to  be  perfect  gases. 
An  imperfect  gas  or  vapor  is  one  that  is  readily  liquefied  under 
ordinary  conditions;  thus,  steam,  which  is  a  vapor  of  water 
(and  a  gas)  condenses  when  allowed  to  expand,  and  is,  therefore, 


§4 


PNEUMATICS 


45 


an  imperfect  gas.  In  what  follows,  air  will  be  taken  as  an  example 
of  a  perfect  gas,  and  the  laws  relating  to  air  will  apply  to  all 
other  perfect  gases. 

74.  The  Atmosphere. — The  earth  is  entirely  surrounded  by  an 
invisible  gaseous  envelope  called  the  atmosphere,  and  the 
mixture  of  gases  composing  the  atmosphere  is  called  air.  The 
principal  gases  composing  the  air  are  nitrogen  and  oxygen; 
in  addition  to  these,  there  is  also  a  small  percentage  of  carbon 
dioxide,  water  vapor,  and  minute  quantities  of  several  other  gases. 
For  practical  purposes,  the  thickness  of  the  gaseous  envelope 
(depth  of  the  atmosphere)  is  about  50  miles,  but  the  actual  depth 
may  be  several  hundred  miles.  The  density  is  greatest  at  the 
surface  of  the  earth,  but  greater,  of  course,  at  the  bottom  of  a  deep 
well  or  shaft,  and  decreases  as  the  distance  above  sea  level 
increases. 

75.  The  atmosphere  exerts  a  pressure  due  to  its  weight  on 
everything  it  touches;  that  this  is  the  case  can  be  shown  in 
many  ways,  but  most  strikingly  in  the  following  manner:  Take 
a  glass  tube  that  is  about  40  inches 
long  and  closed  at  one  end.  Fill  the 
tube  with  mercury,  place  the  hand 
over  the  open  end,  and  keeping  it 
there,  invert  the  tube;  the  mercury 
will  be  felt  pressing  against  the  hand. 
Still  keeping  the  end  of  the  tube 
covered,  insert  the  hand  and  end  of 
tube  in  a  dish  of  mercury,  and  re- 
move the  hand.  If  the  tube  is  verti- 
cal, it  will  be  found  that  the  distance 
between  the  top  of  the  column  and 
the  top  of  the  mercury  in  the  dish  is 
about  30  inches.  See  Fig.  17.  This 
column  of  mercury  is  supported  by  the  pressure  of  the  atmos- 
phere; because,  the  weight  of  the  mercury  exerts  a  pressure  on 
the  mercury  in  the  dish  that  is  transmitted  in  all  directions,  as 
in  the  case  of  any  other  fluid,  and  the  upward  pressure  would 
raise  the  level  of  the  liquid  (mercury)  in  the  disli  were  it  not  for 
the  pressure  of  the  atmosphere  that  just  counterbalances  it. 

76.  The  space  between  the  top  of  the  mercury  column  and  the 
closed  end  of  the  tube,  which  is  the  difference  between  the  length 


Pig.  17. 


46  ELEMENTS  OF  PHYSICS  §4 

of  the  tube  above  the  level  of  the  mercury  in  the  dish  and  the 
length  (height)  of  the  mercury  column,  is  entirely  empty,  or  in 
other  words,  a  vacuum;  this  is  called  a  Torricellian  vacuum, 
after  Torricelli  (1608  — 1647),  who  first  performed  this  experiment 
and  first  showed  that  the  air  had  weight.  If  the  pressure  on  the 
surface  of  the  mercury  in  the  dish  be  increased,  the  column  of 
mercury  in  the  tube  will  rise,  and  if  the  pressure  be  decreased, 
it  will  fall ;  this  shows  conclusively,  that  the  column  is  supported 
by  the  pressure  on  top  of  the  mercury  in  the  dish.  If  the  tube  be 
inclined,  as  shown  in  Fig.  17,  so  that  the  vertical  distance  be- 
tween the  top  of  the  tube  and  the  level  of  the  mercury  in  the  dish  is 
less,  the  mercury  rises  in  the  tube  until  it  fills  it  when  this  vertical 
distance  equals  about  30  inches.  The  top  of  the  mercury  in  A B 
and  the  top  in  the  tube  B'  are  then  in  the  same  horizontal  line. 

77.  The  specific  gravity  of  mercury  is  13.6  at  32°F.,  and  the 
weight  of  1  cu.  in.  =  .49111  lb.;  hence,  a  column  of  mercury 
1  in.  high  exerts  a  pressure  of  .49111  lb.  per  sq.  in.  The 
height  of  the  mercury  column  due  to  the  atmospheric  pressure 
varies  considerably,  being  dependent  on  the  altitude  of  the  place 
above  sea  level,  the  temperature,  the  amount  of  water  vapor 
contained  in  the  air,  etc.,  but  the  standard  value  is  760  milli- 
meters at  0°C,  at  sea  level;  this  corresponds  to  29.9213  inches 
at  32°F.  (=  0°C),  at  sea  level.  The  pressure  exerted  by  the 
atmosphere  is  therefore  29.9213  X  .49111  =  14.6946  lb.  per  sq. 
in.  =  14.6946  X  144  =  2116  lb.  per  sq.  ft.  This  pressure  is 
called  1  atmosphere,  and  whenever  pressures  are  given  in  atmos- 
pheres, they  can  be  changed  into  pounds  per  square  inch  by  multi- 
plying the  number  of  atmospheres  by  14.6946,  which  is  usually 
expressed  as  14.7.  Since  1  lb.  per  sq.  in.  =  703.09456  Kg.  per 
sq.  m.,  1  atmosphere  is  equal  to  703.09456  X  14.6946  =  10,332 
kilograms  per  square  meter.  Since  the  specific  pressure  of  1 
atmosphere  is  2116  lb.  per  sq.  ft.,  this  means  that  everything  on 
the  earth's  surface  is  subjected  to  a  pressure  of  over  a  ton  on 
every  square  foot.  The  reason  that  we  do  not  notice  this 
enormous  pressure  is  that  the  air  within  the  body  has  practi- 
cally the  same  tension  as  the  pressure  of  the  air  outside,  and 
one  counteracts  the  other.  It  becomes  very  apparent,  however, 
on  top  of  a  high  mountain  or  when  in  a  balloon  or  airplane  at 
a  great  height  from  the  earth ;  the  tension  of  the  air  within  the 
body  is  then  greater  than  the  pressure  outside,  and  may  result 
in  the  bursting  of  small  blood  vessels. 


§4  PNEUMATICS  47 

78.  Partial  Vacuum. — A  perfect  vacuum  is  a  closed  space 
that  contains  nothing  that  can  exert  a  pressure  on  the  walls  that 
enclose  the  space.  For  practical  purposes,  the  Torricellian 
vacuum  mentioned  in  Art.  76  is  a  perfect  vacuum;  it  is  not  ac- 
tually so,  because  there  is  a  very  small  amount  of  air  present, 
probably  held  in  suspension  within  the  mercury,  but  the  pressure 
it  exerts  is  so  small  that  it  can  hardly  be  detected.  The  space 
within  a  carbon-filament  electric-light  bulb  is  very  nearly  a  per- 
fect vacuum. 

If,  in  Fig.  17,  a  connection  be  made  between  the  top  of  tube  B 
and  a  receptacle  containing  air  (or  other  gas),  and  a  little  air  be 
admitted  above  the  mercury  column,  the  pressure  of  this  air  will 
counterbalance  a  like  weight  of  mercury;  the  column  will  then 
shorten  by  this  amount.  Thus,  suppose  the  pressure  of  the  air 
enclosed  above  the  mercury  is  2  lb.  per  sq.  in.;  this  is  equiva- 
lent to  2  -T-  .49111  =  4.0724  inches  of  mercury,  and  the  height  of 
the  mercury  column  will  then  be  29.9213  -  4.0724  =  25.8489 
in.,  say  25.85  in.  The  space  above  the  mercury  column  is  then 
called  a  partial  vacuum.  Partial  vacuums  are  nearly  always 
measured  by  the  number  of  inches  of  mercury  that  will  be  sus- 
tained by  the  difference  between  the  pressure  within  the  partial 
vacuum  and  the  pressure  of  the  atmosphere.  For  instance,  the 
partial  vacuum  just  referred  to  would  be  called  a  vacuum  of 
25.85  inches.  It  may  be  remarked  that  vacuum  gauges  are 
graduated  in  inches  of  mercury  and  read  from  0  to  30  inches. 

79.  For  rough  calculations,  the  weight  of  a  cubic  inch  of  mer- 
cury is  taken  as  ^  pound  and  the  pressure  of  the  atmosphere  as 
15  pounds  per  square  inch;  consequently,  a  vacuum  of,  say,  19 
in.  represents  (roughly)  a  mercury  column  of  19  X  .5  =  19  -5-  2 
=  9.5  lb.,  and  the  pressure  within  the  partial  vacuum  is  15  —  9.5 
=  5.5  lb.  per  sq.  in. 

If  exact  results  are  desired,  use  the  following  formula  in  which 
m  =  inches  of  vacuum  and  p  =  pressure  in  pounds  per  square 
inch  within  the  partial  vacuum : 

p  =  14.6946  -  .49111m 

Applying  this  formula  to  the  preceding  case,  p  =  14.6946 
-  .49111  X  19  =  5.3635  lb.  per  sq.  in. 

80.  Examples  of  partial  vacuums  are  very  numerous.  When 
soda  water  is  sucked  through  a  straw,  the  air  in  the  mouth  is 
drawn  into  the  lungs,  a  partial  vacuum  is  created  in  the  mouth, 


48 


ELEMENTS  OF  PHYSICS 


§4 


and  the  pressure  of  the  atmosphere  forces  the  fluid  up  through  the 
straw.  The  conmion  suction  pump,  the  siphon,  etc.  all  operate 
by  reason  of  a  partial  vacuum.  What  are  called  vacuum  pumps 
are  used  to  exhaust  air  from  the  suction  boxes  underneath  the 
wire  of  a  paper  machine,  with  the  result  that  as  the  pulp  on  the 
wire  passes,  the  water  is  forced  through  by  the  fact  that  the 
atmospheric  pressure  above  it  is  greater  than  the  pressure  in  the 
partial  vacuum  below  it,  and  the  water  is  said  to  be  sucked 
through  the  wire. 

81.  Action  of  a  Suction  Pump. — The  manner  in  which  a  suction 
pump  acts   will   be   clear  after   considering  the  diagrammatic 


(a) 


(b) 


Fig.  18. 


sections  in  Fig.  IS.  D  is  the  pump  barrel  in  which  works  a 
piston  P,  the  piston  being  forced  up  and  down  by  pushing  and 
pulling  on  the  piston  rod  R.  The  pipe  A  connects  the  pump 
barrel  to  the  water  supply,  and  where  it  joins  the  barrel  is  a  valve 
o,  which  is  lifted  as  shown  when  the  water  is  entering  the  pump. 
The  piston  also  has  valves  v'  and  v".  When  the  piston  is  moving 
upward,  as  shown  at  (a),  it  leaves  a  partial  vacuum  in  the  space 


§4  PNEUMATICS  49 

B  below  it;  the  atmospheric  pressure  on  top  of  the  open  water 

causes  the  water  to  rise  and  fill  this  partial  vacuum  and  follow  t  he 
piston  upward.  When  the  piston  stops,  the  water  in  B  tends  to 
fall  downward,  but  the  weight  of  the  water  acts  as  a  pressure  on 
top  of  the  valve  v  and  closes  it,  thus  keeping  the  water  in  B. 
When  the  piston  moves  down,  as  shown  in  (b),  valves  v'  and  v" 
lift,  the  water  in  B  passes  through  the  openings  in  the  piston, 
permitting  the  piston  to  move  down.  When  the  piston  again 
moves  upward,  the  weight  of  the  water  above  it  closes  the  valves 
v'  and  v",  and  the  water  in  B'  is  lifted  with  the  piston  and  dis- 
charged through  C;  at  the  same  time,  a  partial  vacuum  is  created 
in  B,  the  atmospheric  pressure  forces  the  water  up  the  pipe  A, 
lifts  valve  v,  and  follows  the  piston,  as  before. 

82.  Height  of  Lift  of  a  Suction  Pump. — Theoretically,  the 
maximum  height  of  "suction"  is  equal  to  the  height  of  a  column 
of  water  that  will  just  balance  a  perfect  vacuum.  In  the  case  of 
mercury,  this  is  29.9213  in.,  and  as  the  specific  gravity  of  mercury 
is  13.6,  the  height  of  a  water  that  will  produce  the  same  specific 
pressure  is  29.9213  X  13.6  =  406.93  in.  =  33.91  ft.  Conse- 
quently, if  the  piston  left  a  perfect  vacuum  behind  it,  and  there 
were  no  friction  or  other  resistances  to  the  movement  of  the  water, 
a  suction  pump  would  lift  water  about  34  ft.  In  practice,  from 
28  to  30  ft.  is  the  greatest  height  that  water  can  be  raised  with  a 
suction  pump.  It  might  be  thought  that  since  the  specific  weight 
of  hot  water  is  considerably  less  than  that  of  cold  water,  hot 
water  could  be  raised  to  a  greater  height  than  cold  water;  such, 
however,  is  not  the  case,  the  reason  being  that  hot  water  gives 
off  vapor  (steam),  and  the  tension  of  this  vapor  creates  a  pres- 
sure that  tends  to  destroy  the  vacuum.  In  fact,  hot  water  can- 
not be  raised  to  as  great  a  height  as  cold  water;  the  hotter  the 
water  the  shorter  the  lift. 

83.  Work  Required  to  Operate  a  Pump. — Since  work  is  equal 
to  force  (pressure)  multiplied  by  the  distance  through  which  it 
acts,  the  work  required  to  operate  a  pump  is  equal  to  the  distance 
moved  by  the  piston  in  one  stroke  multiplied  by  the  lifting  force 
applied  to  the  piston  and  this  product  multiplied  by  the  number 
of  lifting  strokes.  A  stroke  is  the  distance  passed  through  by  the 
piston  between  its  lowest  and  highest  positions.  The  force 
required  to  lift  the  piston  is  the  weight  of  a  column  of  water 
having  the  same  diameter  as  the  piston  and  whose  height  is  equal 


50  ELEMENTS  OF  PHYSICS  §4 

to  the  difference  of  level  between  the  point  of  discharge  and  the 
level  of  the  water  in  the  reservoir  or  other  source  of  supply,  indi- 
cated by  h  in  (a),  Fig.  18.  That  this  is  true  is  readily  seen. 
Thus,  when  the  piston  moves  up,  it  lifts  a  column  of  water 
above  it  equal  to  the  weight  of  a  column  of  water  having  a 
diameter  equal  to  that  of  the  piston  and  a  length  equal  to  the 
stroke ;  at  the  same  time  the  piston  creates  a  partial  vacuum  that 
causes  the  water  to  follow  the  piston,  and  the  amount  of  water 
that  thus  follows  the  piston  is  equal  to  the  amount  discharged  in 
one  stroke.  The  effect  is  exactly  the  same  as  though  the  water 
above  the  piston  had  been  lifted  the  entire  distance  h.  Letting 
h  =  height  of  lift  in  feet,  d  —  diameter  of  piston  in  inches, 
s  =  stroke  in  inches,  the  volume  of  water  lifted  per  stroke  is  ,7854d2s 
cu.  in.  and  the  weight  is  .7854d2s  X  gfo  =  .028362d2s  pounds; 
this  multiplied  by  h  in  feet  is  the  work  done  during  one  lifting 
stroke.    Letting  w  =  work  in  foot-pounds  for  one  lifting  stroke, 

w  =  .028362« 

Example. — Suppose  that  when  the  piston  is  at  the  upper  end  of  its  stroke, 
the  distance  between  the  bottom  of  the  piston  and  the  level  of  water  supply- 
is  23  ft.  and  that  the  distance  between  the  bottom  of  the  piston  and  the 
point  of  discharge  is  57  ft.  If  the  diameter  of  the  piston  is  8  in.  and  the 
stroke  is  12  inches,  what  is  the  work  done  during  one  lifting  stroke? 

Solution.— The  total  lift  of  the  water  is  23  +  57  =  80  ft.  =  h;  the  water 
on  top  of  the  piston  must  be  raised  12  in.,  the  length  of  the  stroke;  that  is,  a 
column  of  water  57  ft.  high  must  be  raised  12  in.  The  height  of  suction  is 
23  ft.  and  the  amount  of  water  discharged  during  one  stroke  must  be  raised 
through  this  height  also.  Consequently,  the  water  discharged  must  be 
raised  through  a  total  height  of  57  +  23  =  80  ft.  Substituting  in  the  form- 
ula the  values  given, 

w  =  .028362  X  82  X  12  X  80  =  1742.6-  foot-pounds.     Ans. 

The  actual  work  done  would  be  considerably  greater  on  account  of  fric- 
tion and  other  resistances. 

84.  The  Siphon. — A  siphon  is  essentially  a  bent  pipe  or  tube 
that  conveys  a  liquid  from  a  point  of  higher  level  to  one  of  lower 
level,  the  highest  point  of  the  pipe  being  higher  than  the  water 
level  of  the  supply.  Thus,  referring  to  Fig.  19,  the  bent  pipe 
A  BCD  connects  vessels  M  and  N,  and  the  highest  point  B  of  the 
pipe  is  higher  than  A,  the  level  of  the  liquid  in  M.  Assuming 
that  the  liquid  is  water,  water  will  flow  from  M  through  the  pipe 
to  N  as  long  as  the  water  level  in  M  is  higher  than  in  N.  A 
siphon  will  not  start  itself,  but  when  the  part  .4  BC  of  the  siphon 
pipe  is  filled  with  water,  water  will  flow  from  M  to  N  until  the 


§4 


PNEUMATICS 


51 


vessel  M  is  emptied  or  the  water  level  is  the  same  in  both  vessels. 
The  reason  that  the  water  rises  in  the  part  A B  of  the  siphon  is 
that  when  the  water  falls  in  the  part  BD,  it  leaves  a  partial 
vacuum  behind  it,  and  the  atmospheric  pressure  on  the  water  at 
A  forces  the  water  up  the  part  AB.  Evidently,  the  height  EB 
=  h'  must  not  exceed  34  ft. ;  in  practice,  it  is  not  advisable  to 
have  it  exceed  28  ft.  The  height  FE  =  h  is  the  difference  of  level 
of  the  liquid  in  the  two  vessels.  If  h  is  in  feet,  h  X  £$  is  the 
specific  pressure  in  pounds  per  square  inch  that  urges  the  water 
from  M  to  N;  because  BF  X  %%  =  specific  pressure  at  D 
due  to  the  water  in  BD,  and  this  is  decreased  by  the  partial 
vacuum  represented  by  EB  X  £f,  thus  leaving  an  active 
specific  pressure  of  (BF  —  BE)  {}  =  h  X  f§. 


i 

\E 

bsS 

li 

i 

M 

i 

C 

^HK| 

h 

" 

rci 

V 

i  -  ■-■■■-  ■_ 

.V 


Fig.  19. 


If  the  pipe  is  a  small  rubber  tube,  the  siphon  may  be  started 
when  the  height  h!  is  not  too  great  by  placing  the  end  D  in  the 
mouth  and  sucking  out  the  air.  If  the  water  level  in  both  vessels 
is  in  the  same  horizontal  plane,  the  siphon  will  not  work,  since 
there  will  then  be  no  specific  pressure  due  to  the  difference  in 
the  water  levels.  The  action  of  a  siphon  may  be  interrupted 
and  resumed  by  means  of  a  valve  or  clamp  on  the  delivery  pipe. 

85.  Barometers. — The  word  barometer  is  derived  from  the 
Greek,  and  literally  means  weight  measurer;  it  is  an  instrument 
used  for  measuring  the  pressure  of  the  atmosphere.  There  are 
two  forms  in  common  use:  the  mercurial  barometer  and  the 
aneroid    barometer. 


52 


ELEMENTS  OF  PHYSICS 


H 


A  common  form  of  mercurial  barometer  is  shown  in  Fig.  20. 
It  consists  of  a  glass  tube  A,  closed  at  the  upper  end,  with  the 
lower  end,  which  is  open,  inserted  in  a  cup  of  mer- 
cury C,  the  arrangement  being  similar  to  that  shown 
in  Fig.  17.  The  tube  and  cup  are  attached  to  a 
wooden  frame  F,  to  which  is  also  attached  a  scale  S 
and  an  accurate  thermometer  T.  The  scale  S  car- 
ries a  vernier,  by  means  of  which  (in  the  instrument 
shown)  readings  may  be  taken  to  }{ o  mm.,  or  about 
.004  in.  "Whenever  a  reading  of  the  barometer  is 
taken,  a  reading  of  the  attached  thermometer  is 
also  taken,  and  then,  by  means  of  tables  or  by  cal- 
culation, if  accuracy  is  desired,  a  correction  is  made 
of  the  reading  to  reduce  it  to  0°  C.  (=  32°  F.). 
Another  correction  is  made  to  reduce  the  reading  to 
sea  level,  and  a  thud  correction  is  made  for  capil- 
larity. When  all  these  corrections  have  been  made, 
the  final  result  is  called  the  barometric  pressure, 
and  it  represents  the  pressure  of  the  atmosphere, 
under  standard  conditions,  at  the  time  the  reading 
was  taken,  the  pressure  being  in  millimeters  or 
inches  of  mercury,  according  to  how  the  scale  is 
graduated. 


Fig. 


Fig.  21. 


86.  Aneroid  barometers  are  made  in  many  forms,  some  being  of 
the  shape  and  size  of  a  watch ;  they  contain  no  liquid.     A  common 


§4  PNEUMATICS  53 

form  is  shown  in  Fig.  21;  it  is  a  metal  box,  circular  in  shape,  air- 
tight,and  carryinga  glassface  and  dial,  the  latter  being  graduated 
in  inches  and  tenths.  The  air  is  exhausted  from  the  box,  and  the 
back  is  so  constructed  and  fastened  to  a  set  of  multiplying  levers 
that  a  very  slight  movement  of  the  back  will  result  in  a  large 
movement  of  the  hand  that  passes  over  the  face  of  the  dial.  An 
increase  in  the  air  pressure  presses  the  back  in  slightly  and  moves 
the  hand ;  a  decrease  in  the  air  pressure  causes  the  hand  to  move 
the  other  way.  These  barometers  are  compensated  for  tempera- 
ture; and  they  are  frequently  so  made  that  they  can  be  adjusted 
for  any  particular  altitude  so  as  to  read  for  sea  level.  When 
well  made,  they  are  very  accurate,  and  some  will  show  a  differ- 
ence in  pressure  for  a  vertical  height  of  only  a  few  feet,  say  be- 
tween the  floor  and  the  top  of  a  table. 

87.  Barometers  are  employed  for  three  purposes:  to  show  the 
altitude  of  a  place  above  sea  level  or  the  difference  of  altitude 
between  two  places;  to  show  the  barometric  pressure  at  any 
place  at  some  particular  time;  and  to  indicate  changes  in  the 
weather.  In  connection  with  the  latter  purpose,  it  is  to  be  re- 
membered that  air  containing  water  vapor  is  less  dense  than  dry 
air;  hence,  if  several  observations  at  regular  intervals  throughout 
the  day  show  a  falling  barometer,  the  moisture  content  (humid- 
ity) of  the  air  is  increasing,  and  if  this  continues,  a  storm  is 
probable.  A  rising  barometer,  on  the  contrary,  shows  that  the 
air  is  becoming  dryer,  and  fair  weather  is  probable. 

88.  Absolute  Pressure.— Suppose  a  large  book,  say  a  diction- 
ary, weighing  20  pounds  to  rest  on  top  of  a  table,  the  pressure  on 
the  table  due  to  the  dictionary  is  20  pounds.  But  there  is  also  a 
pressure  on  top  of  the  table  (and  the  book)  due  to  the  atmosphere. 
The  sum  of  the  two  pressures  is  the  total  pressure  on  the  top  of 
the  table,  and  is  called  the  absolute  pressure.  The  absolute 
pressure,  then,  is  the  sum  of  the  total  external  pressures,  which 
may  be  called  the  apparent  -pressure,  on  any  surface  and  the 
atmospheric  pressure  on  that  surface.  Ordinarily,  only  the 
apparent  pressures  need  be  considered,  since  the  atmospheric 
pressure  in  any  direction  is  balanced  by  an  equal  (or  practically 
equal)  pressure  in  the  opposite  direction.  Thus,  in  the  case  of  the 
table,  the  downward  pressure  on  the  upper  surface  of  the  top  is 
balanced  by  an  upward  pressure  on  the  lower  surface,  and  the 
pressures  on  the  edges  are  also  balanced.     Therefore,  insofar  as 


54  ELEMENTS  OF  PHYSICS  §4 

any  movement  of  the  table  is  concerned,  the  atmospheric  pressure 
has  no  effect. 

Absolute  pressures  begin  at  0,  called  zero  absolute,  which  is 
the  pressure  in  a  perfect  vacuum,  and  they  increase  upward  in- 
definitely. There  can  be  no  such  thing  as  a  negative  absolute 
pressure. 

89.  Pressure  Gauges. — Pressures  of  liquids  and  gases  are 
usually  measured  by  means  of  instruments  called  gauges,  and 
pressures  recorded  on  such  instruments  are  called  gauge  pressures. 
The  dial  of  a  gauge  is  always  so  graduated  that  the  zero  point 
corresponds  to  the  pressure  of  the  atmosphere  only  on  the  fluid; 
in  other  words,  to  the  pressure  on  the  fluid  when  acted  on  only 
by  the  atmosphere.  Consequently,  the  absolute  pressure  is 
always  equal  to  the  gauge  pressure  plus  the  barometric  pressure. 
If  the  barometric  pressure  is  not  known,  add  14.7  to  the  gauge 
pressure,  if  the  pressure  is  recorded  on  the  dial  in  pounds  per 
square  inch.  In  what  are  called  vacuum  gauges,  which  record 
pressures  below  0,  gauge  pressure,  the  scale  readings  are  in 
inches  of  mercury,  and  when  converted  into  pounds  per  square 
inch,  pressures  in  the  partial  vacuum  (see  Art.  79),  are  absolute 
pressures.  Gauges  are  also  made  to  read  in  metric  units,  as 
millimeters  of  mercury  or  pressure  per  square  centimeter. 

Pressures  on  solids  and  liquids  are  nearly  always  taken  as 
pressures  above  the  atmosphere,  i.  e.,  as  gauge  pressures;  in  the 
case  of  gases,  however,  as  will  presently  be  shown,  it  is  frequently 
necessary  to  use  absolute  pressures. 


EXAMPLES 

(1)  Convert  (a)  100°F.  into  degrees  centigrade;  (b)  —  40°C.  into  degrees 
F.;  (c)  1265°C.  into  degrees  Fahrenheit.  f  (a)  37J^°C. 

Ans.    I  (b)   -40°F. 
{  (c)  2309°F. 

(2)  What  is  (a)  the  tension  of  the  steam  in  a  condenser  when  the  vacuum 
is  24.4  in.?  (b)  A  tension  of  12.5  lb.  per  sq.  in.  is  equivalent  to  how  many 
inches  of  vacuum?  .  f  (a)  2.712  lb.  per  sq.  in. 

\  (6)  4.47  in.  of  vacuum. 

(3)  The  diameter  of  a  pump  piston  is  6  in.,  its  stroke  is  9  in.,  and  the 
total  height  of  lift  is  52  ft.  9  in.;  what  is  the  work  done  during  one  stroke? 

Ans.  521.5  ft.-lb. 


ELEMENTS  OF  PHYSICS 

(PART  1) 


EXAMINATION  QUESTIONS 

(1)  If  a  body  be  weighed  on  a  spring  scale  in  Boston  and  also 
in  Baltimore,  in  which  place  will  it  weigh  the  more  and  why? 

(2)  A  certain  body  weighs  1  ton  =  2000  pounds  on  a  beam 
scale  that  has  been  standardized  to  g  =  980.665;  how  much  will 
it  weigh  on  a  spring  scale  in  San  Diego  that  has  been  standardized 
to  the  value  of  g  at  that  place,  if  g  =  32.1373? 

Ans.  1997.71  lb. 

(3)  A  body  weighs  500  pounds  and  has  a  velocity  of  40  ft.  per 
sec.  Due  to  the  action  of  a  steady  force,  the  velocity  is  increased 
to  75  ft.  per  sec.  in  3  seconds.  What  is  (a)  the  acceleration? 
(b)  what  force  is  required  to  produce  this  change  in  velocity? 

(a)   llf  ft.  per  sec.2 


1  (6)   181.4  lb. 

(4)  A  stone  block  (ashlar)  is  8  in.  high,  14  in.  wide,  and  9  ft. 
long.  If  its  specific  gravity  is  2.5  what  is  (a)  the  weight  of  the 
block?  what  is  the  density  of  the  stone?       .         {  (a)   1092    lb. 

AnS-    I  (6)  4.85 

(5)  The  plunger  of  a  hydraulic  jack  on  which  the  load  is  sup- 
ported is  4  in.  in  diameter;  the  piston  that  forces  the  water  or 
other  liquid  into  the  jack  has  a  diameter  of  \  in.;  what  load  will 
the  jack  lift,  if  the  pressure  on  the  piston  is  55  lbs? 

Ans.  3520  lb. 

(6)  Referring  to  the  last  example,  if  the  piston  moves  3|  in. 
during  each  stroke  (a)  how  many  strokes  are  required  to  raise 
the  load  2f  in.?  (b)  how  much  work  would  be  done  in  doing  this? 

(a)  50  strokes. 


1  (b)  80.67    foot-lb. 
(7)  A  specimen  of  a  certain  alloy  weighs  2  lb.  3  oz.  in  air  and 
1  lb.  13f  oz.  in  water;  what  is  the  specific  gravity  of  the  alloy? 

Ans.  8.667. 
§4  55 


56  ELEMENTS  OF  PHYSICS  §4 

(8)  Referring  to  the  last  example,  what  was  the  volume  of  the 
specimen  in  cubic  inches?  Ans.  9.09  cu.  in. 

(9)  How  many  degrees  Fahrenheit  are  equivalent  (a)  to 
—120°  C?  to  625°C.?  .         j  (a)  — 184°F. 

AnS-    1(6)   1157°F. 

(10)  Comparatively  low  pressures  are  sometimes  measured 
by  means  of  an  instrument  called  a  manometer,  which  records 
the  pressure  in  inches  of  mercury.  If  the  manometer  reads 
45.83  in.  of  mercury,  what  is  (a)  the  equivalent  pressure  in 
pounds  per  square  inch?  (b)  in  atmospheres? 

Ang     j  (a)  22.51  lb.  per  sq.  in. 
\  (6)   1.531  atmospheres. 

(11)  Convert  (a)  1859°F.  and  — 210°F.  into  degrees  centigrade. 

/  (a)  1015°C. 
A7lS-    I  (6)  — 135°C. 

(12)  What  is  the  pressure  in  a  condenser  when  the  vacuum 
gauge  reads  22  in.?  Ans.  3.89  lb.  per  sq.  in. 

(13)  The  plunger  of  a  steam  pump  is  2\  in.  in  diameter  and  its 
stroke  is  5  in. ;  the  height  of  suction  is  17  ft. ;  the  level  of  discharge 
is  26  ft.  above  the  pump;  and  the  pump  makes  240  working 
strokes  per  minute.  How  much  work  must  be  done  per  minute 
by  the  pump?  Am.  9147  ft.-lb. 

(14)  The  specific  gravity,  of  a  certain  liquid  is  1.52  and  of 
another  liquid  .977;  (a)  what  would  be  the  hydrometer  reading 
of  the  first  liquid  in  degrees  Beaume?  (6)  of  the  second  liquid? 

f  (a)  49.6°Be. 
S-    1(6)   13.3°Be. 

(15)  The  hydrometer  reading  for  a  certain  liquid  that  is 
lighter  than  water  is  26°  Be.,  and  for  another  liquid  that  is 
heavier  than  water  the  reading  is  also  26°  Be.;  what  is  (a)  the 
specific  gravity  of  the  first  liquid?  (6)  of  the  second  liquid? 

j(a)     .8974. 
AUS'    1(6)    1.2185. 


ELEMENTS  OF  PHYSICS 

(PART  2) 


PNEUMATICS 

(Continued) 


PROPERTIES  OF  PERFECT  GASES 

90.  Boyle's  Law. — While  a  theoretically  perfect  gas  does  not 
exist,  air,  hydrogen,  nitrogen,  oxygen,  and  many  other  gases 
may  be  considered  as  perfect  gases  in  ordinary  practical  applica- 
tions, and  Boyle's  law  (also  called  Mariotte's  law)  may  be  used 
in  all  such  cases.     This  law  may  be  stated  as  follows: 

The  temperature  remaining  the  same,  the  volume  of  any  perfect 
gas  varies  inversely  as  its  absolute  pressure. 

Let  p\  and  vi  be  the  absolute  pressure  and  the  volume  of  some 
perfect  gas,  say  air,  at  a  temperature  of  t°;  suppose  the  pressure 
is  changed  to  p2,  then,  the  temperature  being  still  t°,  the  volume 
v2  may  be  found  from  the  proportion, 

P\  :  P2  =  v2  :  Vi 
in  accordance  with  Boyle's  law.     From  this  proportion, 

PlVi  =  P2V2; 
and,  in  general, 

P)Vi  =  P2V0  =  p3v3  =  etc.  =  k 

in  which  k  is  a  constant. 

Example. — If  the  volume  of  air  in  an  air  compressor  is  1.82  cu.  ft.  at  atmos- 
pheric pressure  (14.7  lb.  per  sq.  in.)  and  it  is  compressed  to  38  lb.  per  sq. 
in.,  gauge,  what  is  its  volume,  the  temperature  being  the  same  in  both 
cases? 

Solution. — The  absolute  pressure  before  compression  is  14.7  and  after 
compression,  38  +  14.7  =  52.7  lb.   per  sq.   in.     Substituting  in  the  above 

14  7  X  1  82 
formula,  14.7  X  1.82  =  52.7  X  v2,  or  n:  =     '-" L-_9  =^~~    =  -5077  ~  cu-   ft- 

Ans. 

91. — In  accordance  with  Boyle's  law,  if  the  pressure  be  doubled, 
the  volume  will  be  halved,  or  if  the  pressure  be  halved  (by  ex- 
§4  57 


58  ELEMENTS  OF  PHYSICS  §4 

pansion  of  the  gas),  the  volume  will  be  doubled.  In  general,  if 
the  pressure  be  increased  (or  decreased)  n  times,  the  volume  will 

be  decreased  (or  increased)  —  times.     This  will  be  evident  from 

n 

the  formula;  because,  if  p2  =  npi,  PiV\  =  npiVi  and  v2  =  -V\. 

Example. — Suppose  air  at  a  gauge  pressure  of  45  pounds  per  sq.  in.  and 
occupying  a  space  of  984  cu.  in.  is  allowed  to  expand  to  4  times  its  volume. 
If  the  barometric  pressure  is  30.106  in.,  what  is  the  pressure,  gauge,  after 
expansion,  the  temperature  remaining  the  same? 

Solution. — The  absolute  pressure  before  expansion  is  45  +  14.7  =  59.7 
lb.  per  sq.  in.  Since  the  volume  increased  4  times,  the  pressure  after  ex- 
pansion is  59.7  -4-  4  =  14.925  lb.  per  sq.  in.,  absolute,  and  the  gauge  pres- 
sure is  14.925  —  14.7  =  .225  lb.  per  sq.  in.     Ans. 

The  same  result  might  have  been  obtained  by  applying  the  formula  of 

the  last  article.     Thus,  the  volume  after  expansion  was  984  X  4  =  3936  cu. 

59  7  X  984 
in.;  hence,  59.7  X  984  =  p2  X  3936,  and  jh  =  ^3935 —  =  14.925. 

92.  The  density  and  specific  weight  (weight  of  a  unit  of  volume) 
vary  directly  as  the  pressure  and  inversely  as  the  volume,  the 
temperature  remaining  the  same.  This  is  a  direct  consequence  of 
Boyle's  law,  as  a  little  consideration  will  show.  Suppose  the 
volume  of  .256  pound  of  air  at  a  temperature  of  t°  and  a  pressure 
of  14.7  lb.  per  sq.  in.,  absolute,  is  3.2  cu.  ft.  The  density  and 
specific  weight  of  a  gas  are  taken  as  equal  numerically;  and  the 

.256 
density  in  this  case  is  -w-~  =  .08.     Now,  if  this  air  be  allowed  to 

expand  to,  say,  3  times  its  original  volume,  it  will  occupy  a  space 

of  3.2  X  3  =  9.6  cu.  ft.,  and  its  density  will  be  —^  =  .02%, 

(which  is  ^3  the  original  density.     Or,  if  the  pressure  be  increased 

n  times,  the  volume  will  be     times  as  large,  and  the  density  will 

be  n  times  as  great.  Let  u>\  and  Wi  be  the  specific  weights 
(densities)  of  the  gas,  Vi  and  v%  the  corresponding  volumes,  and 
pi  and  p*  the  corresponding  pressures,  absolute;  then, 

W1V1  =  W0V2  (1) 

and  W1P2  =  w»px  (2) 

Example. — The  weight  of  1  liter  of  air  at  0°  C.  and  a  pressure  of  1  atmos- 
phere is  1.2929  grams;  what  is  the  weight  of  15  cu.  m.  when  the  pressure  is 
4.6  atmospheres,  both  pressures  being  absolute? 

Solution. — Applying  formula  (2),  u\  =  1.2929,  Pi  =  1,  p>  =  4.6,  and 
1.2929  X  4.6  =  Wi  XI,  or  w*  =  1.2929  X  4.6  =  5.94734  grams  per  liter. 


§4 


PNEUMATICS 


59 


Since  1  liter  =  volume  of  1  cu.  dm.,  and  1  cu.  m.  =  1000  cu.  dm.,  15  cu.  m. 
at  0°  C.  and  under  a  pressure  of  4.6  atmospheres  weigh 

15  X  1000  X  5.94734  =  89210  g.  =  89.21  Kg.     Ans. 

93.  Gay-Lussac's  Law. — Referring  to  Fig.  22,  C  is  a  cylinder, 
closed  at  one  end  and  open  at  the  other,  and  containing  a  tightly 
fitting  piston  P  on  which  is  laid  a  weight  W.  The  space  beneath 
the  piston  is  filled  with  air  under  atmospheric  pressure.  When 
the  piston  is  released,  the  weight  of  the  piston  and  of  the  load  W 
increases  the  specific  pressure  on  the  confined  air;  the  piston 
moves  downward,  compressing  the  air  until  its  tension  exactly 
equals  the  specific  pressure  of  the  atmosphere  plus  the  specific 
pressure  due  to  the  piston  and 
weight  W.  Suppose  the  area  of 
the  piston  is  25  sq.  in.,  its  weight 
is  10  pounds,  and  weight  of  W 
is  20  pounds.  The  specific  pres- 
sure due  to  weight  of  P  and  W  is 

10  +  20 

per  sq. 


>2^^=i=z====^-_-_-_-|F0 


25 


=  1.2    lb. 


in. 


lii_ 


W 


H 


'  .nil  Ihiiiiiiii 


J 


Fig.  22. 


and  the  tension  of  the  confined 
air,  assuming  the  barometric 
pressure  to  be  14.7  pounds  per 
sq.  in.,  is  14.7  4-  1.2  =  15.9 
lb,  per  sq.  in.,  absolute.  If, 
therefore,  the  volume  under- 
neath the  piston  before  it  was  released  be  represented  by  v\  and 
after  release  by  v2,  the  change  in  volume  may  be  represented  by 
the  distance  that  the  piston  moves  down.  For,  if  h  is  the 
distance  between  the  bottom  of  the  cylinder  and  the  bottom 
of  the  piston,  before  release  (indicated  by  the  dotted  lines)  and 
h'  is  the  position  after  release,  vi  =  25  X  h  and  v2  =  25  X  h',  and 

»i        25h        h       _,        ,  .  .       ,  .  .    , 

—  =  SET/  =  n-      therefore,  any  change  in  the  tension  of  the 

confined  air  will  produce  a  change  in  volume  that  may  be  meas- 
ured by — will  be  proportional  to — the  distance  moved  through 
by  the  piston  up  or  down.  It  is  assumed,  for  convenience,  that 
the  piston  moves  without  friction  and  that  any  change  in  the 
tension  of  the  air,  however  slight  it  may  be,  will  cause  a  move- 
ment of  the  piston. 

Suppose  the  temperature  of  the  confined  air  is  0°  C.  (32°  F.), 
as  shown  by  a  thermometer  placed  within  the  cylinder  and  that 


60  ELEMENTS  OF  PHYSICS  §4 

heat  is  applied  to  the  cylinder.  It  will  be  found  that  for  every 
degree  rise  in  temperature  centigrade,  indicated  by  the  ther- 
mometer, the  piston  rises  Tr^rd  of  the  distance  hf;  that  is,  for  every 
degree  rise  in  temperature,  the  volume  increases  Tr^rd  part. 
Xote  that  the  tension  of  the  air  is  not  changed,  since  the  specific 
pressure  is  the  same  as  before.  The  heat  has  increased  the 
vibratory  motion  of  the  molecules  of  air  and  caused  the  air  to 
expand.  If  the  cylinder  be  cooled,  so  that  the  temperature  of  the 
air  is  less  than  0°  C,  the  piston  falls  ?ktid  of  the  distance  h'  for 
every  degree  fall  in  temperature.     Therefore, 

When  the  pressure  remains  the  same,  the  volume  varies  directly  as 
the  temperature. 

Xow  suppose  the  piston  be  fixed  so  it  cannot  move;  the  volume 
then  remains  constant.  Attach  a  gauge  to  the  cylinder  to  meas- 
ure the  tension  of  the  confined  air.  On  heating  the  cylinder,  it 
will  be  found  that  for  ever}'  degree  rise  in  temperature,  the  pres- 
sure (tension)  of  the  air,  as  shown  by  the  gauge,  increases  -jf^d; 
and  for  every  decrease  of  1°  C.  in  temperature,  the  pressure  falls 
2+5  d.     Therefore, 

When  the  volume  remains  the  same,  the  pressure  varies  directly  as 
the  temperature. 

These  two  statements  may  be  expressed  in  the  following 
general  law: 

The  pressure  and  volume  of  a  confined  gas  vary  directly  as  the 
temperature,  when  either  volume  or  pressure,  respectively,  is  constant. 

This  is  commonly  called  Gay-Lussac's  law;  it  is  also  frequently 
called  Charles"  law,  being  named  after  two  French  scientists  who 
discovered  it. 

94.  Since  1°  C.  =  f  =  1.8°  F.,  and  273  X  1.8  =  491.4,  it 
follows  that  for  every  degree  rise  in  temperature  indicated  by  a 
Fahrenheit  thermometer,  the  volume  or  pressure  will  increase 

.p.  ,  =  .002035th,  and  vice  versa,  according  as  the  pressure  or 

volume  is  constant. 

Let  v  =  the  volume  of  the  air,  p  =  tension  of  the  air,  t 
=  number  of  degrees  change  in  temperature,  and  c  =  the  in- 
crease in  volume  or  pressure  due  to  l°rise  in  temperature;  then, 
when  the  pressure  is  constant, 

vi  —  v  +  cvt  =  v(l  +  ct);  (1) 

and  when  the  volume  is  constant, 

Pi  =  p  +  cpt  =  p{\  +  ct).  (2) 


§4  PNEUMATICS  61 

If  a  centigrade  thermometer  is  used,  c  =  -jls  =  -003663;  if  a 
Fahrenheit  thermometer  is  used,  c  =  .002035.  Note  that  if  the 
temperature  be  increased  273°  C.  =  491.4°  F.,  the  volume  or 
pressure  will  then  be  doubled,  since  the  quantity  in  parenthesis, 
1  +  ct,  then  becomes  1  +  Tl  j  X  273  =  1  +  1  =  2. 

If  the  temperature  falls,  formulas  (1)  and  (2)  become 
vi  =  v  —  cvt  =  v(l  —  ct),  (3) 

and 

Pi  =  p  —  cpt  =  p(l  -  ct).  (4) 

Example  1.— If  the  temperature  of  3.42  cu.  m.  of  air  is  raised  86°  C.  under 
constant  pressure,  what  is  the  volume? 

Solution. — Applying  formula  (1),  v  =  3.42  and  t  =  86;  whence,  vi 
=  3.42(1  +  .003663  X  86)  =  3.42  X  1.31502  =  4.4974  cu.  m.     Ans. 

Example  2. — A  certain  amount  of  air  is  cooled  at  constant  volume  from 
32°  F.  and  a  tension  of  45.46  lb.  per  sq.  in.  absolute,  to  -92°  F.;  what  is  the 
tension  after  cooling? 

Solution.— Applying  formula  (4),  p  =  45.46  and  t  =  32°  -  (-92°) 
=  124°;  whence,  Pl  =  45.46(1  -  .002035  X  124)  =  45.46  X  .74766  =  33.989 
lb.  per  sq.  in.     Ans. 

95.  Absolute  Temperature. — Referring  to  formulas  (3)  and 
(4)  of  the  last  article,  t  =  t0  —  h,  in  which  t0  is  always  the  tem- 
perature at  0°  C.  or  32°  F.  Strictly  speaking,  t  in  formulas  (1) 
and  (2)  is  always  equal  to  h  -  t0.  With  this  understood,  it  will 
be  seen  that  if  t  =  273°  C.  in  formula  (3),  vi  =  v{l  -TfiX 
273)  =  0,  that  is,  the  volume  disappears;  this,  of  course  is  im- 
possible, since  air  is  matter,  and  matter  cannot  be  destroyed. 
If,  however,  273  be  substituted  for  t  in  formula  (4),  pr  =  0,  that 
is,  the  air  has  no  tension — its  molecules  have  ceased  to  vibrate. 
For  this  reason,  a  temperature  of  273°  C.  below  0°  C.  is  called 
absolute  zero  of  temperature.  Since  temperature  is  a  measure 
of  the  vibrations  of  the  molecules  of  a  body,  just  as  weight  is  a 
measure  of  the  earth's  attraction  on  a  body,  it  is  assumed  that 
at  absolute  zero  =  —273°  C,  there  is  no  molecular  movement. 

The  temperature  indicated  by  the  thermometer  is  usually 
indicated  by  t,  and  the  temperature  above  0°,  absolute,  called  the 
absolute  temperature,  is  indicated  by  T.  The  absolute  tempera- 
ture centigrade  is  therefore  equal  to  T  =  t  +  273,  usually 
written  273  +  t.  Thus,  the  absolute  temperature  of  the  boiling 
point  'of  water  at  sea  level,  with  the  barometer  at  760  mm.,  is 
273  +'  100  =  373°  C. 

Since  273°  C.  =  491.4°  F.,  and  0°  C.  =  32°  F. ,  absolute  zero 
Fahrenheit  =  -491.4°  +  32°  =  -459.4°;  that  is,  absolute  zero 


62  ELEMENTS  OF  PHYSICS  §4 

on  the  Fahrenheit  scale  is  459.4°  below  0  of  the  Fahrenheit  ther- 
mometer. To  find  the  absolute  temperature  on  the  Fahrenheit 
scale,  add  459.4  to  the  reading  of  the  Fahrenheit  thermometer; 
thus,  the  boiling  point  of  water  has  an  absolute  temperature  of 
459.4  4-  212  =  671.4°  F.  =  671.4  X  f  =  373°  C. 

If  the  temperature  is  below  zero  on  either  scale,  the  absolute 
temperature  will  be  found  in  exactly  the  same  manner.  Thus, 
25°  below  zero,  centigrade  =  -  25°  C,  and  273°  4-  (  -  25°) 
=  273°  -  25°  =  248°  C,  absolute;  and  -25°  F.  =  459.4°  -  25° 
=  434.4°  F.,  absolute. 

96.  The    quantity   in   parenthesis  in  formulas  (1)  — (4),  Art. 

cxa  u  .  1  ,  t         273  +  t         T  , 

94,  may  be  written  1  4-  ^1  =  1  +  ™=  =  =  ^;  and 

t         273  —  t         T 
1  —  270  =       g-q —  =  275.     Consequently,    formulas    (1)    and 

(3)  may  be  written 

V273)   -  CVT>  <D 


/  T 
vi  =  pl- 


ane! formulas  (2)  and  (4)  may  be  written 

Pi  =  pQ  =  cpT,  (2) 

the  value  of  c  being  the  same  as  in  the  preceding  formulas,  and  T 
being  the  absolute  temperature  after  heating  or  cooling.  These 
two  formulas  assume  that  the  original  temperature  of  the  gas  is 
273°  C.  =  491.4°  F.  If  the  original  temperature  is  T9  and  the 
final  temperature  7\,  then 

t!  :  vo  =  Tx  :  T0 
and  pi  :  p0  =  7\  :  T0 

from  which 


and 


I'oTi  .  . 

vi  =  -fp-  (3) 

j  0 

Pi  =  hp~  (4) 

J  0 


Applying  formula  (2)  to  the  second  example  of  Art.  94,  T0 
=  459.4  +  32  =  491.4,  7\  =  459.4  -  92  =  367.4,  pQ  =  45.46  and 

45.46  X  367.4 
Pi  =  401^4 — "~  =  33.989  lb.   per  sq.  in.,  the  same  result  as 

was  previously  obtained, 


§4  PNEUMATICS  63 

Example. — If  a  constant  volume  of  gas  is  heated  from  60°  F.  to  340°  F., 
and  the  final  tension  is  36  lb.  per  sq.  in.,  gauge,  what  was  the  original 
tension? 

Solution.— T0  =  459.4  +  60  =  519.4,     T,  =  459.4  +  340  =  799.4,     pi 

799  4 
=  36  +  14.7  =  50.7.      Applying    formula    (4),    50.7  =  ^  X  Po,  or  p0 

50  7  X  519  4 
=      '  7QCJ4    '-  =  32.9  lb.   per  sq.  in.,  absolute  =  32.9  -  14.7  =  18.2  lb. 

per  sq.  in.,  gauge.     Ans. 

It  is  useless  to  obtain  more  than  3  significant  figures  when  the  barometric 
pressure  is  assumed  as  14.7  pounds  per  sq.  in. 

97.  Combined  Law  of  Boyle  and  Gay-Lussac. — Boyle's  law 
and  Gay-Lussac 's  law  may  be  combined  into  a  single  law,  as 
follows: 

The  product  of  the  pressure  and  volume  of  any  perfect  gas  varies 
directly    as   the   absolute    temperature. 

Let  po,  v0,  and  T0  be  the  original  pressure,  volume,  and 
temperature,  ph  Vi,  and  7\  the  pressure,  volume,  and  temperature 
after    a    change;    then 

*%  -  IT  -  *£  "  etc-  =  *        (i) 

in  which  R  is  a  constant  for  any  particular  gas,  but  has  different 
values  for  different  gases.     To  find  the  value  of  R  for  air,  write 

PoVo         „ 
1  0 

Here  p0  is  the  absolute  specific  pressure  =  14.6946  lb.  per 
sq.  in.  at  sea  level,  with  the  barometer  at  29.9213  in.,  and  the 
thermometer  at  32°  F.,  for  which  T„  =  32°  +  459.4°  =  491.4°. 
To  fix  the  value  of  v0,  note  that  the  volume  of  1  pound  of  air 

(the  specific  volume)   under  the    above    conditions  is    nRft7  > 

since  1  cu.  ft.  of  air  under  these  conditions  weighs   .08071  lb. 
Substituting    these    values    in    the    last    equation, 
14  6946  X  1 
491.4  X  .08071  =  Ri  °r  R  =  -370506'  say  -37051 

Consequently,    for    air,    ~  =  .37051,    or 

pv  =  .37051  T  (2) 

Formula  (1)  can  be  applied  only  when  the  weight  of  the  air  is  1 
pound;  to  make  it  applicable  to  any  weight  of  air  w  lb.,  mul- 
tiply   the    right-hand    member    of    the    equation    by   w,  and 

pv  =  .37051  wT  (3) 


64  ELEMENTS  OF  PHYSK  -  §4 

To  find  the  value  of  R  for  any  gas  other  than  air,  divide.  .  370506 
by  the  specific  gravity  of  the  gas.  Thus,  the  specific  gravity  of 
hydrogen  is  .06926.  and  the  value  of  R  for  hydrogen  is 
.370506  -s-  .06926  =  5.3495:  the  specific  gravity  of  oxygen  is 
1.10535.  and  the  value  of  R  for  oxygen  is  .370506  -5-  1.10535 
=  .33519. 

In  formulas  (2)  and  (3  .  p  is  the  pressure  (absolute.;  in  pounds 
per  square  inch,  p  is  the  volume  in  cubic  feet.  T  is  the  absolute 
temperature.  Fahrenheit,  and  w  is  the  weight  of  the  air  in  pounds. 
In  formula  (1),  v%,  n,  etc..  may  be  expressed  in  any  units — cubic 
feet,  cubic  meters,  etc. 

Example  1. — What  is  the  weight  of  18,000  cubic  feet  of  air  at  a  tempera- 
ture of  60°  F.  and  a  tension  of  26.3  lb.  per  sq.  in.,  gauge? 

Soi/mox.— The  absolute  pressure  is  14.7  —  26.3  =  44  lb.  per  sq.  in. 
and  the  absolute  temperature  is  459.4  —  60  =  519.4°.  Substituting  in 
formula  (3),  44  X  1S000  =  .37051  X  if  X  519.4;  whence. 

44  X  1SO00  .„_,.         . 

w  =        n  .  -    .   =  411o.olb.     Am. 

.oiOol  X  ol9.4 

Example  2. — The  volume  of  the  cylinder  of  an  air  compressor  is  28,276 
cu.  in.  If  the  reading  of  the  barometer,  corrected,  is  30. 1  inches,  and  the 
temperature  of  the  air  is  56;  F..  it  can  be  shown  that  if  the  air  be  compressed 
to  one-fourth  its  volume  without  losing  any  heat  by  radiation  or  otherwise, 
the  pressure  (tension,  will  be  104.4  lb.  per  sq.  in.,  absolute;  what  will  be 
its  temperature? 

Solution. — Since  the  volume,  pressure,  and  temperature  all  change, 
apply  formula  (1).  Here  p:  =  the  pressure  of  the  atmospheric  air  =  30.1 
X  .49111  =  14.782  lb.  per  sq.  in.,  r0  =  28,276  cu.  in.,  T.  =  459.4  -  56 
=  515.4,  p-  =  104.4  lb.  per  sq.  in.,   and  r;  =  28,276  X  H  =  7069  cu.  in. 

_       .  .    14.782  X  28276        104.4  X  7069    _  ,  .  . 

Therefore,   bv  formula   (l;.  .,  -  . = ^ .  from  which. 

515.4  7; 

r,  =  1Q4;1  yj^^S^—    =  910=  absolute    =  910   -  459.4  =  450.6°  F. 
14. /82  X  28- 

.4  ns. 

Example  3. — What  is  the  specific  volume  of  atmospheric  air  when  the 
barometer  stands  at  29.652  in.  and  the  temperature  is  100:  F.? 

Solution. — The  specific  volume  is  the  volume  of  1  pound  in  cubic  feet; 
hence,  applying  formula  (2).  p  =  29.652  X  .49111  =  14.562  lb.  per  sq. 
in..    T  =  459.4  —  100  =  559.4.    and    14.562  X  r  =  .37051  X  559.4:    from 

.37051  X  559.4 

which.  ■  =  ,  .  _grt =  14.233  cu.  ft.     Ans. 

14.O0J 

98.  Mixture  of  Gases. — Suppose  two  liquids  of  different  den- 
sities to  be  mixed  and  that  the  mixture  is  allowed  to  stand.  If 
there  be  no  chemical  action  between  the  liquids,  it  will  be  found 
that  after  a  time  the  liquids  have  separated,  the  lighter  floating 
on  the  heavier.     The  case  is  entirely  different  with  gases,  since 


§4  PNEUMATICS  65 

if  two  vessels  containing  gases  of  different  densities  are  connected 
with  each  other,  it  will  be  found  after  a  time  thai  thetwo  gases  have 
mixed,  the  density  of  the  mixture  in  both  vessels  being  the  same; 
even  though  the  vessel  containing  the  lighter  gas  be  placed  over 
the  other,  the  same  result  will  be  obtained — the  density  and 
quality  of  the  mixture  of  the  gases  in  both  vessels  will  be  the  same. 
This  is  an  extremely  useful  property  of  gases;  otherwise  no  life 
of  any  kind  could  exist.  As  an  example  of  this  fact,  it  must  be 
borne  in  mind  that  when  anything  is  burned  or  when  any  animal 
or  vegetable  matter  decays,  carbon  dioxide  gas  is  given  off,  it  be- 
ing a  product  of  combustion  and  decay,  which  is  slow  combustion. 
Since  carbon  dioxide  is  more  than  1)4  times  as  heavy  as  air,  it 
follows  that  if  gases  separated  in  accordance  with  their  densities, 
as  in  the  case  of  liquids,  the  carbon  dioxide  would  form  a  layer 
covering  the  entire  earth,  and  would  kill  all  forms  of  animal  life. 

Rule. — If  two  or  more  gases  are  mixed,  the  'product  of  the  pressure 
and  volume  of  the  mixture  is  equal  to  the  sum  of  the  products  of  the 
pressures  and  volumes  of  the  gases,  provided  there  is  no  chemical 
action  and  the  temperature  of  the  gases  is  the  same. 

Let  P  and  V  be  the  pressure  and  volume  of  the  mixture,  pi 
Pi,  Pz,  etc.,  the  pressures  of  the  gases,_and  vi,  v2,  v3,  etc.,  the  corres- 
ponding volumes;  then 

PV  =  p\vx  +  poVo  +  P3V3  +  etc. 

Example. — Suppose  two  vessels  to  contain  air  at  the  same  temperature. 
The  volume  of  one  vessel  is  5.72  cu.  ft.  and  the  tension  of  the  air  is  17  lb. 
per  sq.  in.,  gauge;  the  volume  of  the  other  vessel  is  8.36  cu.  ft.  and  the  ten- 
sion of  the  air  is  24  lb.  per  sq.  in.,  gauge;  if  they  are  allowed  to  communi- 
cate with  a  third  vessel  (empty)  having  a  volume  of  3.48  cu.  ft.,  what  will  be 
the  pressure  of  the  mixture? 

Solution.— The  volume  of  the  mixture  is  5.72  +  8.36  4-  3.48  =  17.56 
cu.  ft.,  since  if  both  vessels  communicate  with  the  empty  vesssel,  the  air 
can  mix  in  all  three  vessels.  Then,  applying  the  formula,  the  absolute 
pressures   are    14.7  4-  17  =  31.7    and    14.7  +  24  =  38.7,   and    P  X  17.56 

504  856 
=  31.7X5.72  4-38.7  X  8.36  =  504.856,  from  which,  P  =     y,  —     =  28.75 

lb.  per  sq.  in.,  absolute  =  28.75  -  14.7  =  14.05  lb.  per  sq.  in.,  gauge.     Ans. 
If   the  gases  are  alike   and  the   temperatures   are  different, 
then,  letting  T,  7\,  T2,  etc.,  be  the  corresponding  absolute  tem- 
peratures, 

PV  _  P1V1    ,    P2V2   ■    P3V3    1      . 
T    ~    Tx  +    Tz  +    T»  +CTC- 

Example. — In  the  example  of  the  last  article,  suppose  the  temperature 
of  the  air  in  the  smaller  vessel  had  been  95°  F.,  in  the  larger  vessel  50°  F., 
5 


66  ELEMENTS  OF  PHYSICS  §4 

and  the  temperature  of  the  mixture  after  a  time  was  found  to  be  62°  F. ; 
what  was  the  tension  of  the  mixture  at  that  time. 

Solution. — The  absolute  temperatures  are  T  =  459.4  +  62  =  521. 4°, 
Ti  =  459.4  +  95  =  554.4,  and  Tt  =  459.4  +  50  =  509.4.  Applying  the 
,  PX  17.56       31.7  X  5.72    ,   38.7  X  8.36         _olo 

formula,         521  4       =  55^4 H 5Q04 =  -96219>  from  whlch 

^21  4  y    9R219 
P  =         '  17  56  =   28<57   lb'    Per    SQ'    in-'    absolute    =   28-57  _  14-7 

=  13. S7  lb.  per  sq.  in.,  gauge.     Ans. 


EXAMPLES 

(1)  Referring  to  Fig.  22,  if  1.86  cu.  ft.  of  air  under  a  gauge  pressure  of 
51.51b.  per  sq.  in.  be  allowed  to  expand  to  4.28  eu.  ft.  by  decreasing  the 
weight  W,  what  will  be  the  pressure  if  there  is  no  change  in  the  tempera- 
ture? Ans.  13.94  lb.  per  sq.  in.,  gauge. 

(2)  Referring  to  Art.  97,  what  is  the  weight  of  220,000  cu.  ft.  of  hydrogen 
at  75°F.  under  a  pressure  of  1  atmosphere?  Ans.  1130.8  lb. 

Note. — When  stated  in  this  manner,  the  pressure  is  always  understood  to  mean  absolute 
pressure  and  under  standard  conditions  i.e.,  14.6946  lb.  per  sq.  in. 

(3)  A  certain  volume  of  air  has  a  tension  of  21.7  lb.  per  sq.  in.,  gauge, 
and  a  temperature  of  265°F. ;  it  is  cooled  until  the  temperature  is  62°F. ; 
what  is  the  tension?  Ans.   11.93  lb.  per  sq.  in.,  gauge. 

(4)  A  vessel  containing  750  cu.  in.  of  free  air  has  forced  into  it  2120  cu.  in. 
of  air  under  a  pressure  of  52.63  lb.  per  sq.  in.,  gauge.  If  the  tempera- 
ture of  both  volumes  of  air  and  of  the  mixture  is  the  same  and  the  baro- 
metric pressure  is  30.6  in.  of  mercury,  what  is  the  tension  of  the  mixture? 

Ans.   191.25  lb.  per  sq.  in.,  gauge. 

Note. — When  the  barometric  pressure  is  given,  it  must  be  used  instead  of  14.7  or  14.6946 
lb.  per  sq.  in.  in  finding  the  absolute  pressure,  and  vice  versa. 

(5)  If,  in  the  last  example,  the  temperature  of  the  free  air  had  been  50° 
F.,  and  that  of  the  air  in  the  other  vessel  208°,  what  is  the  tension  of  the 
mixture  when  its  temperature  is  92°?         Ans.  158.51  lb.  per  sq.  in.,  gauge. 


HEAT 


NATURE  AND  MEASUREMENT  OF  HEAT 

99.  Nature  of  Heat. — Heat  may  be  defined  as  molecular  energy. 
The  molecules  of  a  body  have  mass,  and  since  they  are  in  rapid 
motion,  they  possess  kinetic  energy,  which  manifests  itself  in  the 
phenomenon  called  heat.  At  the  temperature  of  absolute  zero, 
all  movement  of  the  molecules  ceases,  the  body  posessess  no 
heat,  and  is  absolutely  inert — dead.  For  any  temperature  above 
0°,  absolute,  the  body  has  an  amount  of  heat  that  is  proportional 
to   the   temperature,   the  temperature  being   dependent    upon 


§4  HEAT  07 

the  velocity  of  the  molecules  and  being  a  measure  of  the  amount  of 
heat  contained  in  the  body. 

Heat  is  energy,  and  since  energy  manifests  itself  in  different 
forms,  heat  may  be  changed  into  different  forms  of  energy,  such 
as  mechanical  energy,  chemical  energy,  and  electrical  energy; 
conversely,  these  three  forms  of  energy  may  all  be  converted  into 
heat. 

100.  Quantity  of  Heat. — The  temperature  of  a  body  does  not 
depend  upon  its  mass  or  size.  This  is  evident  from  the  fact  that 
if  an  ordinary  pin  and  a  ten-pound  iron  weight  be  placed  in  a  tub 
of  hot  water,  the  pin,  weight,  and  water  will  all  have  the  same 
temperature  after  a  time.  In  other  words,  temperature  may  be 
regarded  as  a  measure  of  the  specific  molecular  energy  of  a  body; 
it  is  therefore  independent  of  the  size  and  shape  of  the  body. 
At  the  same  time,  it  is  obvious  that  the  iron  weight  contains  a 
greater  amount  of  heat  (kinetic  molecular  energy)  than  the  pin, 
and  it  is  desirable  to  have  some  means  of  measuring  heat  in 
quantity,  and  this  is  accomplished  by  (a)  measuring  the  amount 
of  energy  expended  in  raising  the  temperature,  that  is,  by  chang- 
ing another  form  of  energy  into  heat;  (6)  measuring  the  energy 
produced  by  a  fall  of  temperature,  when  changing  heat  into  some 
other  form  of  energy. 

The  most  obvious  way  of  measuring  heat  is  by  measuring  the 
work  required  to  overcome  friction.  Assuming  that  there  is  no 
wear  or  that  the  wear  is  so  exceedingly  small  that  it  can  be  dis- 
regarded, all  the  work  of  friction  is  converted  into  heat.  Taking 
as  a  standard  for  one  heat  unit  the  work  of  friction  required  to 
heat  one  pound  of  water  from  63°F.  to  64°F.,  it  has  been  found  by 
careful  and  elaborate  tests  that  the  amount  of  work  required  for 
this  purpose  is  very  closely  778  foot-pounds,  which  is  called  the 
mechanical  equivalent  of  heat.  When  the  centigrade  scale  is 
used,  the  work  required  to  raise  1  kilogram  of  water  from  17° 
to  18°C.  (62.6°F.  to  64.4°F.)  is  very  nearly  427  meter-kilograms. 

101.  The  British  thermal  unit  (abbreviation,  B.t.u.)  is  the 
amount  of  heat  required  to  raise  the  temperature  of  one  pound 
of  water  1°  F.  (from  63°  to  64°),  and  its  mechanical  equivalent 
is  778  foot-pounds. 

The  calorie  (abbreviation,  cal.)  is  the  amount  of  heat  required 
to  raise  the  temperature  of  one  kilogram  of  water  1°C.  (from  17° 
to  18°),  and  its  mechanical  equivalent  is  427  m.-kg. 


68  ELEMENTS  OF  PHYSICS  §4 

The  reason  for  specifying  the  temperature  is  that  the  amount 
of  heat  required  to  raise  the  temperature  of  a  given  weight  of 
water  one  degree  is  different  for  different  temperatures. 

Since  1  kg.  =  2 .  204622  lb.  and  1°C.  =  |  =  1.8°F.,  1  caloric  is 
equal  to  2.204622  X  1.8  =  3.96832  B.t.u.  Also,  1  m.-kg. 
=  3.28084  X  2.204622  =  7.2330    ft.-lb.,    and    1    calorie  =  778 

X  ^H?  =  426.8  +,  say  427  m.-kg. 

For  many  purposes,  the  calorie  is  too  large  a  unit  for  convenient 
use,  and  what  is  called  the  small  calorie  or  gram  calorie  is  then 
employed;  this  unit  is,  the  amount  of  heat  required  to  raise  the 
temperature  of  1  gram  of  water  from  17°C.  to  18°C,  and  since 
a  gram  is  l-1000th  of  a  kilogram,  the  gram  calorie  is  l-1000th  of  a 
large  calorie  (kilogram  calorie),  or  1000  gram  calories  =  1  kilo- 
gram calorie.  The  gram  calorie  is  largely  used  by  chemists, 
physicists,  and  scientists  generally,  wmile  the  kilogram  calorie  is 
used  in  engineering  calculations.  Both  units  are  called  calories, 
and  the  reader  must  be  on  his  guard  to  distinguish  which  is 
meant;  but  a  little  familiarity  with  their  use  will  soon  show  which 
is  intended,  since  one  is  1000  times  as  large  as  the  other. 

102.  Heat  Capacity. — The  capacity  of  a  body  for  heat  depends 
upon  the  weight  (strictly  speaking,  the  mass)  of  the  body  and 
upon  the  material  of  which  the  body  is  composed;  for  equal 
weights,  the  capacity  depends  only  on  the  material  composing 
the  body.  It  can  be  shown  by  experiment  that  the  number  of  heat 
units  required  to  raise  the  temperature  of  equal  weights  of  iron, 
lead,  copper,  wood,  etc.  15°,  say,  are  quite  different;  and  when 
these  substances  are  cooled,  the  amount  of  heat  given  off  (which 
is  exactly  the  same  as  that  imparted  in  raising  the  temperature 
the  same  amount  the  body  is  cooled)  is  also  different.  The  more 
heat  required  to  raise  the  temperature  the  greater  is  the  capacity 
of  the  body  for  heat,  and  vice  versa. 

103.  Specific  Heat. — The  specific  heat  of  a  substance  is  the 
amount  of  heat  required  to  raise  the  temperature  of  a  unit 
weight  of  the  substance  1°  divided  by  the  amount  of  heat  re- 
quired to  raise  the  temperature  of  the  same  weight  of  water  1° 
at  some  specified  temperature.  The  specific  heat  of  water  is 
generally  considered  to  be  1,  in  which  case,  the  specific  heat  of 
any  substance  may  be  defined  as  the  number  of  B.t.u.  required 
to  raise  the  temperature  of  the  substance  1°.  If  the  specific 
heat  of  a  substance  be  denoted  by  c,  the  weight  by  w,  and  the 


§4  HEAT  69 

difference  in  temperature  before  and  after  heating  (or  cooling) 
by  t,  the  number  of  heat  units  U  (B.t.u.)  required  to  heat  the 
body  t°  (or  given  off  in  cooling  t°)  is 

U  =  cwt 
For  example,  if  the  specific  heat  of  lead  is  .0315,  how  many 
heat  units  are  required  to  raise  the  temperature  of  25  lbs. 
from58°F.  to  330°F.?  here  t  =  330  -  58  =  272,  w  =  25,  and  c 
=  .0315,  therefore,  the  number  of  heat  units  required  is  U  =  .0315 
X  25  X  272  =  214.2  B.t.u.,  which  is  equivalent  to  214.2  X  778 
=  166,647.6,  say  166,600  foot-pounds  of  work.  When  multi- 
plying by  778  (or  427),  the  mechanical  equivalent,  it  is  useless  to 
retain  more  than  3  or  4  significant  figures  in  the  product;  this 
statement  also  applies  to  multiplying  by  the  specific  heat. 

104.  The  specific  heat  of  most  substances  (if  not  all)  is  not 
constant,  but  varies  somewhat  with  the  temperature  and  with 
the  pressure;  for  this  reason,  results  obtained  by  application  of  the 
formula  in  the  last  article  are  not  strictly  accurate,  but  are 
sufficiently  so  for  most  practical  purposes.  The  following  table 
gives  average  values  of  the  specific  heats  of  some  solids  and  liquids 
that  may  be  used  for  practical  purposes  in  connection  with  engi- 
neering calculations.  Different  authorities  give  slightly  differ- 
ent values.  Some  of  these  values  are  useful  in  pulp  mills,  when 
in  calculating  the  unit  required  to  melt  sulphur,  etc. 

SPECIFIC  HEATS 

SOLIDS 

Aluminum 218  Ice 504     Sulphur 180 

Brass 090  Lead 0315  Tin 055 

Cast  Iron 130  Platinum 0325  Tungsten 034 

Charcoal 203  Steel  (soft) 116     Wrought  Iron 127 

Copper 095  Steel  (hard) 117     Zinc 094 

Gold 032  Silver 056     Glass 199 

LIQUIDS 

Acetic  Acid 51     Hydrochloric  Lead  (melted) 041 

Acid 60 

Alcohol  (pure) 70     Mercury 033  Sulphur  (incited  ) .    .235 

Benzine 43     Sulphuric  Acid.  .    .336  Tin  (melted) 058 

Glycerine 576  Water 1  Turpentine 472 

105.  The  specific  heat  of  gases  has  two  different  values,  ac- 
cording to  whether  the  gas  is  heated  at  constant  volume  or  under 
constant  pressure.  When  heated  at  constant  volume,  all  the 
heat  goes  to  increasing  the  molecular  energy,  that  is,  the  tempera- 


70  ELEMENTS  OF  PHYSICS  §4 

ture;  but  when  heated  under  constant  pressure,  the  gas  expands 
and  does  work,  the  work  being  equal  to  the  increase  in  volume 
multiplied  by  the  tension  of  the  gas.  Consequently,  it  requires 
more  heat  to  raise  the  temperature  of  a  gas  under  constant 
pressure  than  at  constant  volume.  Letting  cp  =  specific  heat  under 
constant  pressure  and  c„  =  specific  heat  at  constant  volume,  the 
following  table  gives  the  specific  heats  of  some  gases. 

SPECIFIC  HEATS 

GASES 
Cp  Cv  Cp  Ct 

Acetylene 270  .024  Hydrogen 3 .  42  2 .  44 

Air." 241  .171  Nitrogen 247  .176 

Carbon  Monoxide. .   .  243  .  172  Oxygen 217  .155 

Carbon  Dioxide 210  .160  Sulphur  Dioxide.      .154  .123 

When  the  pressure  and  volume  both  change,  as  when  a  certain 
volume  of  compressed  air  expands  and  does  work,  the  specific  heat 
does  not  have  either  of  the  two  values  given  above;  the  law  that 
connects  the  pressure  and  volume  under  those  circumstances  is 
quite  complicated,  and  will  not  be  given  here. 

106.  Temperature  of  Mixtures  of  Different  Bodies. — If  two  or 
more  substances  of  the  same  material  or  of  the  same  specific 
heats  and  having  different  temperatures  are  mixed,  the  tempera- 
ture of  the  mixture  is  readily  found.  Suppose,  for  example, 
that  10  pounds  of  water  at  50°  are  mixed  with  10  pounds  of 
water  at  86°;  the  temperature  of  one  must  be  increased  to  the 
temperature  of  the  mixture,  and  the  temperature  of  the  other 
must  be  decreased  to  the  temperature  of  the  mixture;  then,  since 
both  bodies  weigh  the  same,  one  receives  as  much  heat  as  the 
other  loses,  and  the  temperature  of  the  mixture  will  be  midway 

-no      i      n/>o 

between  the  temperatures  of  the  two  bodies,  or ~ =  68°. 

Here  one  body  has  increased  its  temperature  68°  —  50°  =  18°, 
and  the  temperature  of  the  other  body  has  decreased  86° 
-  68°  =  18°. 

Suppose,  now.  that  4  pounds  of  water  at  50°  are  mixed  with 
9  pounds  of  water  at  86°;  one  body  still  receives  as  much  heat  as 
the  other  loses,  but  the  temperature  of  the  mixture  will  not  be  the 
mean  of  the  temperatures  of  the  two  bodies,  because  the  larger 
body  has  a  greater  capacity  for  heat  than  the  smaller  body;  hence, 
any  change  in  the  temperature  of  the  larger  body  will  produce 
a  greater  change  in  the  smaller  body.     Let  x  =  the  temperature 


§4  HEAT  71 

of  the  mixture;  then,  evidently,  (86  —  x)X  9  =  (x  —  50)  X  4, 
since  the  larger  body  gives  up  (86  —  x)  X  9  heat  units  and  the 
smaller  body  receives  the  same  number,  which  must  equal  (x 
—  50)  X  4  heat  units,  the  specific  heat  of  water  being  1.  The 
above  equation  reduces  to  774  —  9x  =  4x  —  200,  which  becomes, 
by  transposition,  lSx  =  974,  or  x  =  74.923°.  This  result 
might  have  been  arrived  at  in  the  following  manner:  let  tm  and 
wm  be  the  temperature  and  weight  of  the  mixture,  ti  and  Wi  the 
temperature  and  weight  of  one  body,  t2  and  w2  the  temperature 
and  weight  of  another  body  of  the  same  material,  etc. ;  then, 
wmtm  =  witi  +  w2t2  +  108*8  +  etc.  (1) 

In  the  present  case,  wm  =  wx  +  w2  =  4  +  9  =  13,  ti  =  50, 
and    t2  =  86;    substituting    in    the    formula, 

13*m  =  4  X  50  +  9  X  86  =  200  +  774  =  974 
from  which,  tm  =  74.923° 

the  same  result  as  before.     It  will  also  be  noted  that  the  same 
operations    are    performed    in    both    cases. 

If  only  the  temperature  of  the  mixture  is  desired,  the  above 
formula  may  be  written. 

,     _  wi  U  +  w2t2  +  w3t3  +  etc.  ,  . 

tm  —  '  ;  ;  ;       I  \A) 

Wi  +  Wo  +  w$  -\-  etc. 

107.  If,  however,  the  bodies  mixed  have  different  specific 
heats,  every  term  in  the  righthand  member  of  the  above  for- 
mula must  be  multiplied  by  the  specific  heat  of  the  corresponding 
body,  since  the  amount  of  heat  given  up  or  received  by  any  body 
is  cwt  (Art.  103).  and  this  is  the  same  for  all  the  bodies,  including 
the    mixture.     In    such    case, 

CiiviU  +  c2w2t2  +  CjWzU  +  etc.  ,Q. 

tm    =   j ; j        7  \<J) 

CiWi  +  c2w2  -j-  C3W3  +  etc. 

Example  1. — Suppose  a  copper  ball  weighing  1.4  pounds  and  having  a 
temperature  of  860°  is  placed  in  a  lead  vessel  weighing  3.8  pounds  and  con- 
taining 7.5  pounds  of  mercury,  the  temperature  of  the  vessel  and  mercury 
being  74°;  after  all  three  bodies  have  arrived  at  the  same  temperature,  with- 
out any  loss  of  heat,  what  is  the  temperature  of  the  mixture? 

Solution. — From  the  table  of  Art.  104.  specific  heat  of  copper  is  .095, 
of  lead  .0315,  and  of  mercury  .033.     Substituting  in  the  above  formula, 

_  .095  X  1.4  X  860  +  .0315  X  '3.8  X  74  +  -033  X  7.5  X  74  ^  . 

.095  X  1.4  +  .0315  X  3.8  +  .033  X  7.5 

It  is  useless  to  use  more  than  three  (or  at  most,  four)  significant  figures 
when  the  constants  (specific  heats)  are  given  to  only  two  or  three  significant 
figures. 

Example  2. — How  many  heat  units  (B.t.u.)  are  required  to  raise  the 
temperature  of  a  cast-iron  ball  weighing  28  lb.  4  oz.  from  28°C.  to  236°C? 


72  ELEMENTS  OF  PHYSICS  §4 

Solution.— Here  t  =  236  -  28  =  208°C.  =  208  X  1.8  =  374.4°F.     Us- 
ing the  formula  of  Art.  103,  c  =  .13,  to  =  28.25  pounds,  and 

17  =  .13  X  28.25  X  374.4  =  1375  B.t.u.     Ans. 

Had  the  result  been  required  in  calories,  change  the  weight  to  kilograms, 
obtaining  28.25  -r-  2.2046  =  12.814  kg.  In  this  case,  t  =  208°C,  and  U 
=  .13  X  12.814  X  208  =  346.5  cal.  Or,  since  1  cal.  =  3.96832  B.t.u., 
1375  -s-  3.96832  =  346.5  cal. 


EXPANSION  OF  BODIES  BY  HEAT 

108.  Kinds  of  Expansion. — When  the  temperature  of  the  body 
is  changed  its  volume  changes,  an  increase  in  temperature  usually 
resulting  in  an  increase  in  volume,  an  important  exception  being 
water,  which  decreases  in  volume  from  32°F.  to  39 °F.  and  then 
increases  for  all  higher  temperatures.  When  the  volume  increases, 
the  body  is  said  to  expand,  and  the  change  in  volume  is  called 
cubic  expansion.  As  the  result  of  cubic  expansion,  the  area  of  the 
surface  and  of  any  section  is  increased,  and  this  change  is  called 
surface  expansion;  likewise,  the  length  of  any  line  is  increased, 
and  this  change  is  called  linear  expansion.  When  a  body  is 
cooled,  the  opposite  effect  is  produced,  that  is,  the  body  contracts, 
which  may  be  called  negative  expansion.  Steel  bridges  generally 
have  one  end  free,  so  they  can  move  in  the  direction  of  the  length 
of  the  bridge,  since  the  bridge  will  be  longer  on  a  hot  day  in  the 
summer  than  on  a  cold  day  in  the  winter;  the  difference  in  length 
may  be  an  inch  or  more,  depending  on  the  length  of  the  bridge 
and  the  difference  in  temperature.  If  both  ends  were  rigidly 
fastened,  this  difference  in  length  would  tend  to  make  the  bridge 
buckle  in  the  summer  and  tend  to  pull  it  apart  in  the  winter. 

109.  Coefficient  of  Expansion. — The  increase  in  length  for  an 
increase  in  temperature  of  1°  between  32°F.  and  33°F.  divided 
by  the  original  length  at  32°  is  called  the  coefficient  of  linear 
expansion ;  similarly,  the  increase  in  area  divided  by  the  original 
area,  and  the  increase  in  volume  divided  by  the  original  volume, 
under  the  same  conditions,  are  called  the  coefficient  of  surface 
expansion  and  coefficient  of  cubic  expansion,  respectively.  In 
the  case  of  solids  and  liquids,  the  coefficient  of  expansion  is  so 
small  that  the  coefficient  of  surface  expansion  is  always  taken  as 
2  times  the  coefficient  of  linear  expansion,  and  the  coefficient  of 
cubic  expansion  is  always  taken  as  3  times  the  coefficient  of  linear 
expansion. 


HEAT 


73 


For  every  degree  (Fahrenheit)  rise  in  temperature,  the  body 
increases  in  length  an  amount  equal  to  the  product  of  the  co- 
efficient (which  is  different  for  different  materials)  and  the  length 
at  32°,  and  when  the  temperature  of  a  body  decreases,  the  con- 
traction in  length  (negative  expansion)  is  equal  to  the  expansion 
of  the  body  for  the  same  rise  in  temperature.  The  coefficient  of 
expansion  varies  somewhat  with  the  temperature,  but  it  is  suffici- 
ently accurate  in  practice  to  multiply  the  length  at  the  original 
temperature  by  the  average  value  of  the  coefficient  and  by  the 
difference  in  temperature  to  find  the  increase  in  length.  Let 
hi,  k-2,  and  A;3  be  the  coefficients  of  linear,  surface  and  cubical 
expansion,  respectively;  let  ti  be  the  original  temperature  and 
t2  the  final  temperature;  let  h  and  l2,  a:  and  a2,  and  Vi  and  v2  be 
the  lengths,  areas,  and  volumes  at  ti  and  t2,  respectively;  then, 
U  =  h  +  kih  (M  -  h)  =  h  [1  +  fei  (U  -  h)]  (1) 

a2  =  ai  [1  +  fca  (t2  -  h)]  (2) 

v2  =  vi  [1  +  h  (t2—ti)]  (3) 

If  the  original  temperature  is  higher  than  the  final  temperature, 
as  when  the  body  cools,  the  quantity  in  parenthesis,  t2 — ti,  will 
be  negative  and  the  final  length,  area,  or  volume  will  be  less,  as  it 
should  be. 

The  following  table  gives  for  temperatures  Fahrenheit  the  coef- 
ficients of  expansion  (average  values)  for  a  number  of  substances. 

COEFFICIENTS  OF  EXPANSION 


A-j 


Aluminum 

Brass 

Copper 

Gold 

Iron  (cast) 

Iron  (wrought) 00000648 

Platinum 

Silver 

Steel 

Tin 

Zinc 

Mercury  (60°F.) 

Alcohol  (ethyl) 

Alcohol  (methyl) 

Gases  (perfect) 


.00001234 

.00002468 

. 00003702 

. 00000957 

.00001914 

.00002871 

. 00000887 

.00001774 

.00002661 

. 00000786 

.00001572 

. 00002352 

. 00000556 

.00001112 

.00001668 

.00000648 

.00001290 

.00001944 

.00000479 

.0000095* 

.00001437 

.00001079 

.00002158 

. 00003237 

. 00000636 

.00001272 

.00001908 

.00001163 

.00002326 

. 00003489 

.00001407 

.00002M4 

.00004221 

. 00003333 

. 0000GG67 

.00010000 

. 000203 

.000407 

.000610 

.000267 

.000533 

.000800 

. 000678 

.001357 

.002035 

The  coefficient  of  cubic  expansion  of  gases  is  1  -r-  491.4  =  .002035. 


74  ELEMENTS  OF  PHYSICS  §4 

The  only  liquids  included  are  mercury  and  alcohol,  because 
these  two  liquids  are  used  in  thermometers  and  because  the 
coefficients  of  expansion  for  liquids  vary  considerably  with  the 
temperature. 

110.  In  machine  shops,  what  are  called  shrinking  fits  are  fre- 
quently used.  For  instance,  the  tires  on  locomotive  drivers  are 
"shrunk  on."  The  part  of  the  driver  on  which  the  tire  is  shrunk 
is  called  the  wheel  center;  this  is  turned  to  a  certain  desired  size, 
and  the  tire  is  bored  a  trifle  smaller.  The  tire  is  then  heated  to 
quite  a  high  temperature  which  causes  it  to  expand  sufficiently 
to  slip  over  the  wheel  center.  When  cooled,  the  tire  contracts 
and  grips  the  wheel  center  with  immense  force.  Crankpins, 
also,  are  frequently  shrunk  on  in  this  manner. 

Example  1. — It  is  desired  to  shrink  a  steel  tire  on  a  locomotive  driver; 
if  the  diameter  of  the  wheel  center  is  73.471  in.  and  the  tire  is  bored  to  a 
diameter  of  73.4  in.,  to  what  temperature  (approximately)  must  the  tire 
be  heated  to  slip  on  the  wheel  center,  assuming  that  the  diameter  when 
heated  is  .0013  in.  larger  than  the  wheel  center? 

Solution*. — Since  the  diameter  is  a  line,  the  case  is  one  of  linear  expansion, 
and  formula  (1)  may  be  used.  Here  h  =  73.4,  h  -  73.471  4-  .0013 
=  73.4723,  h  =  .00000636  (from  table),  and  it  is  desired  to  find  the 
difference  of  temperature  k  —  U.     Solving  formula  (1)  for  U  —  h,  t2   —  U 

=  ~^T  =  73.4  X  .00000S36  =  153°F-  U  the  °riginal  temPerature  of  the 
tire  is,  say  70°F.,  the  temperature  after  heating  is  70  +  155  =  225°F.     Ans. 

Example  2. — A  round  copper  plate  has  a  diameter  of  8.336  in.  at  62°F.; 
what  will  be  the  decrease  in  area  when  the  temperature  is  32°? 

Solution. — The  original  area  is  .7S54  X  8.3362  =  54.5766  sq.  in.  Apply- 
ing formula  (2),  the  area  after  cooling  to  32°  is  (since  k2  for  copper 
=  .00001774)  a,  =  54.5766  [1  +  .00001774^32  -  62)]  =  54.5766  X  .9994678 
—  54.5476  sq.  in.,  and  the  decrease  in  area  is  54.5766  —  54.5476  =  .029 
sq.  in.     Ans. 

The  same  result  might  have  been  obtained  somewhat  more  easily;  thus, 
the  decrease  in  area  is  evidently  equal  to  k%ail  =  ..00001774  X  54.5766 
X  (62  -  32)  =  .029  sq.  in. 

Example  3. — A  silver  vessel  holds  exactly  1  liter  when  the  temperature 
is  20°C;  how  many  cubic  centimeters  will  it  hold  when  the  temperature  is 
100°C? 

Solution. — Formula  (3)  must  be  used  in  this  case,  and  kz  for  silver  is 
.00003237;  the  difference  in  temperature  is  80  -  20  =  60°C.  =  60 
X  1.8  =  108°F.     Therefore,  since  1000  c.c.  =  1  liter  =  ou 

V2  =  1000(1  4-  .00003237  X  10S)  =  1003.5  c.c.  =  1.0035  I.     Ans. 

The  result  obtained  in  the  last  example  shows  how  necessary 
it  is  to  consider  the  temperature  when  accurate  measurements 
of  volume  are  desired. 


§4  HEAT  75 


HEAT  TRANSMISSION 

111.  Heat  may  be  transmitted  or  propagated  from  one  point  or 
place  to  another  point  or  place  in  three  ways:  by  conduction;  by 
convection;  by  radiation. 

112.  Conduction. — If  one  end  of  an  iron  rod  be  heated,  it  will 
be  found  that  after  a  time  the  other  end  is  perceptibly  warmer; 
and  if  the  process  continues,  the  end  not  in  contact  with  the  source 
of  heat  will  become  hot.  The  explanation  is  that  the  molecules 
at  the  heated  end  transmitted  molecular  energy  (heat)  to  the 
adjacent  molecules,  these,  in  turn  transmitted  heat  to  the  mole- 
cules adjacent  to  them,  and  so  on,  until  the  other  end  was  reached. 
The  longer  the  one  end  is  in  contact  with  the  source  of  heat  the 
hotter  will  the  other  end  become;  in  other  words,  heat  is  con- 
ducted from  one  end  to  the  other,  passing  from  molecule  to 
molecule.  Again,  if  a  vessel  containing  a  fluid,  say  water,  be 
placed  in  contact  with  a  fire,  as  a  kettle  of  water  on  a  stove,  the 
vessel  becomes  hot,  and  after  a  time,  the  water  also  becomes  hot. 
In  this  case,  two  different  substances  are  in  contact — the  metal  of 
the  vessel  and  the  water  in  the  vessel — and  the  heat  is  conducted 
through  the  metal,  molecular  energy  is  imparted  to  the  molecules 
in  contact  with  the  metal  and  then  to  the  adjacent  water  molecules, 
until  all  the  water  is  heated. 

Heating  by  conduction  only  is  in  any  case  a  relatively  slow 
process;  but,  nevertheless,  some  materials  conduct  heat  a  great 
deal  faster  than  others.  Those  substances  that  conduct  heat 
best  are  called  good  conductors,  while  those  that  conduct  heat 
poorly  are  called  poor  conductors  or  insulators.  The  metals 
are  all  classed  as  good  conductors,  while  fluids  (except  mercury, 
which  is  a  metal)  are  poor  conductors.  Silver  is  the  best  known 
conductor  of  heat  and  copper  is  next.  All  organic  substances 
are  poor  conductors;  this  enables  trees  and  other  forms  of  vegeta- 
tion to  withstand  sudden  changes  in  the  weather  without  injury, 
since  they  heat  slowly  and  cool  with  equal  slowness.  Moreover, 
the  bark  is  a  poorer  conductor  of  heat  than  the  wood  beneath  it, 
and  this  gives  added  protection.  Rocks,  earth,  soot,  and  all 
loose  materials  transmit  heat  the  more  slowly  as  their  density 
and  uniformity  of  composition  decrease. 

113.  Coefficient  of  Conductivity. — The  coefficient  of  conduc- 
tivity (also  called  the  thermal  conductivity)  is  the  quantity  of 


76 


ELEMENTS  OF  PHYSICS 


H 


heat  that  will  flow  in  one  hour  through  a  plate  one  foot  thick, 
and  having  an  area  of  one  square  foot  when  the  difference  of  tem- 
perature between  the  two  sides  of  the  plate  is  one  degree  Fahrenheit. 
It  may  also  be  expressed  in  calories  per  second  conducted  by 
one  cubic  centimeter,  temperatures  being  in  degrees  centigrade. 
Let  t\  and  U  be  the  temperatures  on  the  two  sides  of  the  plate, 
a  =  area  of  plate  in  square  feet,  b  =  thickness  of  the  plate  in 
feet,  U  =  heat  units  in  B.t.u.  transmitted  per  hour,  and  A; 
=  coefficient  of  conductivity;  then 

U  =  j(h  -  U) 

when  ti  is  the  higher  temperature. 

The  following  table  gives  the  value  of  k  for  a  number  of  different 
materials;  its  value  usually  increases  somewhat  with  the  tempera- 
ture, for  which  reason,  the  temperature  at  which  k  was  deter- 
mined is  also  given.  When  no  temperature  is  given,  it  is  under- 
stood to  be  64°F. 

COEFFICIENTS  OF  CONDUCTIVITY 


Aluminum 116 

Copper 220 

Gold 169 

Iron,  cast  (129°)..   27 

Alcohol  (77°)...  .104 

Benzine  (41°)...  .081 
Petroleum  (55°). 086 
Turpentine    (55°)079 


Asbestos  (100°) 097 

Cotton  (100°) 035 

Cardboard 120 

Cement 170 

Felt 022 

Firebrick 750 

Graphite 2.90 


Metals 
0      Iron,  wrought 34.9 


Silver 244.0 


0      Steel 26.2      Tin 

37.6 

0      Lead 19.8      Zinc 

64.1 

6      Mercury  (32°) 3.6      Brass  (63°)..  . 

63.0 

Fluids 

Water  (63°) 032      Carbon  Dioxide 

(32°) 

. .0079 

Air  (32°) 0126      Hydrogen  (32°) .  . . 

.    0775 

Ammonia  (32°) ...  01 1 1      Nitrogen  (32°) 

.    00S5 

Carbon  Monoxide. 0121      Oxvgen  (32°) 

.0136 

(32°) 

Insulating  Materials 

Silk  (100°) ...  028      Pulverized  Cork  (100°) .  .  026 
Wool  (100°) . .  027      Infusorial    Earth  (100°) .  039 

Miscellaneous 

Linen 050  Rubber 100 

Mica 440  Sand 031 

Mineral  Wool 035  Sawdust 037 

Paper 075  Vulcanite 210 

Porcelain 600  Wood  Ashes 041 


114.  An  inspection  of  the  table  will  show  what  poor  conductors 
the  fluids  are  as  compared  with  the  metals.     Thus,  the  value  of  k 


§4  HEAT  77 

for  steel  is  26.2  ■*■  .0126  =  2000  times  that  for  air,  and  the  value  of 
k  for  copper  is  220  +  .0126  =  17,460  times  that  for  air.  A  layer 
of  air  is  therefore  a  very  good  protection  against  loss  of  heat; 
hence,  when  making  a  boiler  setting,  it  is  always  well  to  make  the 
walls  double,  with  an  air  space  between.  Water  is  also  a  very 
poor  conductor  of  heat;  in  fact,  if  a  piece  of  ice  be  placed  a  little 
below  the  top  surface  of  a  vessel  of  water  and  heat  is  applied 
to  the  water  above  the  ice,  the  water  may  be  boiled  without 
melting  the  ice. 

When  steam  is  conveyed  in  pipes  from  point  to  point,  the  pipe 
should  be  covered  with  some  good  non-conducting  heat  material 
— asbestos,  mineral  wool,  wood  ashes,  etc. — to  keep  the  heat 
within  the  pipe.  A  covering  of  any  kind,  provided  there  is  an  air 
layer  (even  a  thin  one)  between  the  covering  and  the  pipe,  will 
afford 'considerable  protection  against  heat  losses.  Most  pipe 
coverings  owe  their  value  to  the  air  cells  they  contain. 

116.  Convection. — If  a  kettle  of  water  be  placed  over  a  hot 
fire,  the  water  soon  becomes  heated  and,  later,  boils.  The  water 
is  heated  all  the  way  through,  and  the  entire  contents  of  the  kettle 
have  the  same  temperature,  practically  speaking.  Very  little  of 
the  heat  imparted  to  the  water  is  due  to  conduction,  since  if  all 
the  heat  were  due  to  this  cause,  it  would  take  a  very  long  time 
to  raise  all  the  water  to  the  boiling  point.  What  happens  is  this : 
The  particles  in  contact  with  the  heated  surface  are  themselves 
heated,  which  causes  them  to  expand;  their  density  decreases 
owing  to  the  increase  in  volume,  and  they  rise  to  the  top,  colder 
and  denser  particles  taking  their  place.  There  is  thus  created 
a  circulation  of  water  particles,  the  colder  ones  constantly  taking 
the  place  of  the  warmer  ones  until  they  are  all  at  practically  the 
same  temperature.  This  process  is  called  convection.  It  is 
evident  that  convection  can  take  place  only  in  fluids,  (liquids  or 
gases).  Without  convection,  the  only  way  that  a  fluid  could 
be  heated  throughout  in  a  reasonable  time  would  be  by  constantly 
stirring  or  otherwise  agitating  it. 

116.  Radiation. — If  one  approaches  a  hot  stove,  or  any  hot 
body,  a  feeling  of  warmth  is  felt,  which  becomes  the  more  intense 
the  closer  the  body  is  brought  to  the  source  of  heat.  This  feeling 
of  warmth  is  not  due  to  the  heating  of  the  surrounding  air, 
because  if  a  screen  of  some  kind,  a  sheet,  board,  metal  plate, 
etc.,  be  placed  between  the  body  and  the  hot  stove,  the  sensation 


78  ELEMENTS  OF  PHYSICS  §4 

of  warmth  immediately  disappears.  Heat  is  thus  transmitted 
or  propagated  through  an  intervening  space  from  one  object 
to  another  without  heating  appreciably  the  air  between  them. 
This  method  of  transmitting  heat  is  called  radiation,  and  the  heat 
thus  transmitted  is  called  radiant  heat.  All  the  heat  received 
from  the  sun  is  radiant  heat,  the  intervening  space  is  about 
93,000,000  miles,  and  the  question  is,  how  is  this  heat  transmitted? 
The  answer  to  this  question  is  a  direct  conclusion  from  the  modern 
theory  of  heat. 

117.  Dynamical  Theory  of  Heat. — The  modern  dynamical 
theory  of  heat  is  embodied  in  the  following  statement  from 
Ganot's  Physics: 

"A  hot  body  is  one  whose  molecules  are  in  a  state  of  vibration. 
The  higher  the  temperature  of  a  body  the  more  rapid  are  these 
vibrations,  and  a  diminution  in  temperature  is  but  a  diminished 
rapidity  of  the  vibrations  of  the  molecules.  The  propagation  of 
heat  through  a  bar  is  due  to  a  gradual  communication  of  this 
vibratory  motion  from  the  heated  part  to  the  rest  of  the  bar.  A 
good  conductor  is  one  which  readily  takes  up  and  transmits  the 
vibratory  motion  from  molecule  to  molecule,  while  a  poor  con- 
ductor is  one  which  takes  up  and  transmits  the  motion  with 
difficulty.  But  even  through  the  best  of  the  conductors,  the 
propagation  of  this  motion  is  comparatively  slow.  How,  then, 
can  be  explained  the  instantaneous  perception  of  heat  when  a 
screen  is  removed  from  a  fire  or  when  a  cloud  drifts  from  the  face 
of  the  sun?  In  this  case,  the  heat  passes  from  one  body  to  an- 
other without  affecting  the  temperature  of  the  medium  which 
transmits  it.  In  order  to  explain  these  phenomena,  it  is  imagined 
that  all  space,  the  space  between  the  planets  and  the  stars,  as  well 
as  the  interstices  in  the  hardest  crystals  and  the  heaviest  metal — 
in  short,  matter  of  any  kind — is  permeated  by  a  medium  having 
the  properties  of  matter  of  infinite  tenuity,  called  ether.  The 
molecules  of  a  heated  body,  being  in  a  state  of  intensely  rapid 
vibration,  communicate  their  motion  to  the  ether  around  them, 
throwing  it  into  a  system  of  waves  which  travel  through  space 
and  pass  from  one  body  to  another  with  the  velocity  of  light 
[about  186,400  miles  per  second].  When  the  undulations  of  the 
ether  reach  a  given  body,  the  motion  is  given  up  to  the  molecules 
of  that  body,  which,  in  their  turn,  begin  to  vibrate;  that  is  the 
body  becomes  heated.     This  process  of  this  motion  through  the 


§4 


HEAT 


79 


ether  is  termed  radiation,  and  what  is  called  a  ray  of  heat  is 
merely  one  series  of  waves  moving  in  a  given  direction." 

118.  Laws  Governing  Radiation  of  Heat. — As  the  result  of 
experiments,  the  following  laws  have  been  found  to  apply  to  heat 
radiation : 

The  intensity  of  heat  radiated  from  a  given  source  varies  (a) 
as  the  temperature  of  the  source;  (b)  inversely  as  the  square  of 
the  distance  from  the  source;  (c)  grows  less  the  greater  the  angle 
between  the  heat  rays  and  a  perpendicular  to  the  surface  heated. 

The  first  statement  is  to  be  expected.  The  second  law  follows 
from  the  fact  that  if  S,  Fig.  23  (a),  be  a  source  of  heat,  and  ABCD 
and  abed  are  parallel  flat 
surfaces  whose  perpendicu- 
lar distances  from  S  are  OS 
and  oS,  the  solid  formed  by 
the  rays  drawn  to  the  ver- 
tices is  a  pyramid,  and 
from  mensuration,  area  of 
abed  :  area  of  ABCD  =  oS2: 
OS2.  But,  the  intensity  of 
the  heat  on  abed  is  greater 
than  on  ABCD,  since  its 
area  is  less;  and  the  inten- 
sities on  the  two  surfaces 
is  inversely  as  their  areas, 
which  confirms  the  law. 
Regarding  the  third  law,  let  S,  Fig.  23  (6),  be  the  source  of 
heat  and  AB  a  flat  surface,  SO  being  a  perpendicular  from  S  to 
A B.  Evidently,  any  heat  ray  from  S  to  AB  that  is  not  per- 
pendicular to  AB  is  longer  than  OS;  the  greater  the  angle  OSA 
or  OSB  the  longer  will  be  the  line  AS  or  BS,  and  the  greater 
the  area  heated  by  the  ray.  Hence,  the  intensity  of  the  heat 
at  A  or  B  will  be  less  than  at  0. 

Radiant  heat  is  transmitted  through  a  vacuum,  because, 
according  to  theory,  the  vacuum  contains  ether,  which  transmits 
the  vibratory  motion. 

119.  If  two  bodies  having  different  temperatures  are  placed  in  a 
closed  space,  both  bodies  radiate  heat  energy;  but  the  hotter  body 
radiates  more  energy  than  the  colder  body,  with  the  result  that 
after  a  time  both  bodies  will  have  the  same  temperature,  since  the 


80  ELEMENTS  OF  PHYSICS  §4 

colder  body  will  absorb  more  heat  than  the  hotter  body.  Even 
after  both  have  reached  the  same  temperature,  there  will  still  be 
an  interchange  of  heat  energy  between  the  two  bodies. 

It  is  clear  that  the  better  conductor  of  heat  a  body  is  the  more 
heat  it  will  radiate.  In  the  case  of  an  uncovered  steam  pipe,  the 
heat  passes  from  the  inner  surface  to  the  outer  by  conduction  and 
then  leaves  the  pipe  by  radiation.  Therefore,  unless  this  result 
is  desired,  as  in  a  steam-  or  hot-water  heating  plant,  the  pipe 
should  be  covered  with  a  non-conducting  material,  to  keep  the 
heat  in  the  pipe. 

The  amount  of  heat  radiated  depends  upon  the  area  of  the 
surface  radiating  the  heat  and  upon  the  material  of  which  it  is 
composed.  Bright  surfaces  radiate  more  heat  than  dull  ones  and 
white  surfaces  more  than  black.  Similarly,  those  surfaces  that 
radiate  the  most  heat  absorb  the  least ;  it  is  for  this  reason  that 
light  colored  clothes  are  worn  in  the  summer  and  dark  colored 
ones  in  the  winter.  Even  if  the  clothes  are  made  of  the  same 
material  and  have  the  same  weight,  dark  colored  clothes  are 
warmer  than  light  colored  ones. 


EXAMPLES 


(1)  How  many  heat  units  are  equivalent  to  1,980,000  foot-pounds  of 
work?    .  Atis.  2545  B.t.u.,   nearly. 

(2)  How  many  calories  are  equivalent  to  (a)  2,655,229  foot-pounds  of 
work?  \b)  to  1.980.000  foot-pounds  of  work?  Am.    )  860.0  cal. 

\  641.3  cal. 

(3)  A  piece  of  iron  having  a  temperature  of  1040CF.  is  placed  in  a  copper 
vessel  weighing  2  lb.  2  oz.  and  containing  12  lb.  4  oz.  of  water.  The 
weight  of  the  iron  (wrought  iron)  is  1  lb  6  oz.  and  the  temperature  of  the 
water  and  vessel  is  88°.  Assuming  that  there  is  no  loss  by  radiation,  what 
is  the  temperature  of  the  mixture?  Ans.  101. 16T.,  nearly. 

(4)  A  number  of  steel  rails  are  welded  together  until  their  length  is  3000 
ft.;  what  will  be  the  difference  in  length  between  summer  and  winter,  if 
the  range  of  temperature  is  from  110°F.  to  —  10°F.?     Aim.  15.5  in.,  nearly. 

(5)  An  aluminum  vessel  when  filled  to  a  certain  mark  holds  exactly 
1  U.  B    -         231  cu.  in.)  at  62°F.;  how  much  will  it  hold  at  32°F.? 

Ans.  230.742  cu.  in.,  nearly. 

(6)  To  find  the  temperature  of  the  hot  gases  escaping  from  a  blast  fur- 
nace, a  platinum  ball  weighing  }o  lb.  is  placed  in  them;  it  is  then  placed 
in  a  brass  vessel  weighing  10  oz.  and  containing  3  lb.  of  water  at  a  tempera- 
ture of  603F.  The  temperature  of  the  mixture  being  72.4°F.,  what  is  the 
temperature  of  the  gases?  Ans.  2404°.  say  2400°F. 


§4  HEAT  81 

CHANGE  OF  STATE 

120.  Latent  Heat  of  Fusion. — Suppose  a  piece  of  ice  weighing 
1  pound  and  having  the  form  of  a  right  cylinder  with  a  circular 
base  be  placed  in  a  cylindrical  vessel  of  the  same  diameter;  sup- 
pose further  that  the  temperature  of  the  ice  is  32°F.  and  that 
the  pressure  of  the  atmosphere  is  14.7  lb.  per  sq.  in.  If  heat 
be  applied  to  the  vessel,  the  ice  will  gradually  melt  and  become 
water.  But,  so  long  as  there  is  any  ice  left,  the  temperature  does 
not  change;  it  remains  at  32°.  What  becomes  of  the  heat? 
The  ice  does  not  expand;  in  fact,  after  it  is  all  melted,  the  volume 
occupied  by  the  water  is  less  than  the  original  volume  of  the  ice. 
According  to  the  dynamical  theory  of  heat,  the  effect  of  applying 
the  heat  to  the  ice  was  to  overcome  the  attraction  of  cohesion 
to  such  an  extent  that  the  solid  became  a  liquid.  This  heat  is 
called  latent  heat;  it  cannot  be  measured  with  a  thermometer. 
The  heat  that  affects  the  thermometer  is  called  sensible  heat, 
because  it  is  apparent  to  the  senses. 

As  the  result  of  carefully  conducted  experiments,  it  has  been 
found  that  it  requires  144  B.t.u.  of  heat  to  change  1  pound  of  ice 
at  32°  to  water  at  32°,  the  pressure  being  14.7  lb.  per  sq.  in. 
It  has  also  been  experimentally  demonstrated  that  before  water 
at  32°  can  be  changed  into  ice,  144  B.t.u.  of  heat  must  be  re- 
moved from  every  lb.,  the  pressure  being  14.7  lb.  per  sq.  in. 
This  value,  144  B.t.u.,  is  called  the  latent  heat  of  fusion  of  ice. 

The  foregoing  explains  why  water  freezes  (changes  to  ice)  so 
slowly;  it  is  necessary  to  remove  a  large  amount  of  heat  before  the 
liquid  can  become  a  solid.  It  also  explains  why  ice  floats  in  the 
water  in  which  it  was  formed;  since  the  water  expands  in  freezing, 
the  density  of  the  ice  is  less  than  water,  thus  causing  the  ice  to 
float.  Moreover,  after  the  ice  has  become  liquid  (water),  the 
water  continues  to  contract  in  volume — become  denser — until 
a  temperature  of  39°  is  reached;  hence,  the  ice  floats  highest  when 
the  temperature  of  the  water  is  39°F. ;  this  temperature  is  called 
the  temperature  of  maximum  density. 

The  student  will  understand  that  heat  values  can  be  expressed 
in  metric  units  (calories)  by  using  grams,  centimeters  and  degrees 
centigrade.  The  conversion  of  B.t.u.  to  calories,  and  vice  versa, 
is  explained  in  Art.  101. 

121.  Latent  Heat  of  Vaporization. — Assuming  that  the  tem- 
perature is  39°  F.  and  that  the  pressure  is  kept  constant  at  14.7 


82  ELEMENTS  OF  PHYSICS  §4 

lb.  per  sq.  in.,  further  application  of  heat  causes  the  tempera- 
ture of  the  water  to  rise  and  also  causes  the  water  to  expand 
slightly,  thus  increasing  its  specific  volume  and  decreasing  its 
density.  This  action  continues  until  a  temperature  of  212°F.  is 
reached,  when,  if  the  pressure  still  remains  the  same,  the  water 
begins  to  vaporize,  the  temperature  remaining  212°  until  all  the 
water  has  been  converted  into  vapor  (steam).  There  is,  how- 
ever, a  great  increase  in  volume.  The  heat  required  to  change 
the  water  into  steam  of  the  same  temperature  and  pressure  is 
called  the  latent  heat  of  vaporization,  and  when  the  pressure  is 
14.7  lb.  per  sq.  in.,  the  temperature  is  212°,  and  the  number 
of  heat  units  required  is  970.4  B.t.u.  per  pound  of  water.  When 
steam  condenses  and  becomes  water  at  this  temperature,  the 
same  number  of  B.t.u.  is  given  off  or  must  be  removed  from  the 
steam.  Consequently,  if  1  pound  of  ice  and  o  pounds  of  water  at 
32°  are  mixed  with  1  pound  of  steam  at  212°,  the  pressure  being 
14.7  lb.  per  sq.  in.,  the  temperature  of  the  mixture  may  be 
found  approximately  as  follows:  When  the  steam  condenses,  it 
gives  up  970.4  heat  units,  which  go  to  heat  the  water  and  ice, 
and  becomes  1  pound  of  water  at  212°.  When  the  ice  melts,  it 
takes  144  heat  units  to  change  it  into  1  pound  of  water  at  32°, 
and  this  must  be  subtracted  from  the  heat  units  in  the  mixture. 
The  mixture  thus  consists  of  1  +  5  =  6  pounds  of  water  at  32°, 
1  pound  of  water  at  212°,  970.4  B.t.u.,  and  144  negative 
B.t.u.,  the  total  weight  of  the  mixture  being  6+1  =  7  pounds. 

„,,  .    .-       ,        6  X  324-  1  X  212  4-  970.4  -  144 

Ihe  temperature  is  therefore = 

=  175.8°.     The  result  is  approximate,  because  it  has  been  assumed 

that  the  specific  heat  of  water  is  constant  and  equal  to  1;  but,  as 

before  stated,  the  specific  heat  of  water  varies.     However,  the 

result  as  found  is  not  far  out  of  the  way,  and  the  illustration 

serves  to  show  how  the  latent  heats  enter  into  problems  of  this 

Temperature       State    Specific  nature,  such    as   calculating  the 

volume  amount    of    steam    required    to 

-10°C.  =  14°F.     Ice  1.0897  raise  the  temperature  of  liquids 

0°      =  32°        Ice  1.0909  in  a  mill. 

s  :£■  zz  x  „  i2\ hTvr  tabie/venf 

50°  =  122°  Water  1 0i90  nerewi^h    anords    some    idea    of 

100°  =  212°  Water  1.0431  now   ^ie   specific   volume  varies 

100°  =  212°  Steam  26.79      with  the  temperature.     The  vol- 

150°  =  302°  Steam  30.38      Ume  of  1  pound  of  water  at  4°C. 


§4  HEAT  83 

=  39.2°F.  being  taken  as  1,  the  volume  at  212°  will  be  1.0431  and 
the  volume  of  1  pound  of  steam  at  212°  will  be  26.79  cu.  ft. 
the  pressure  being  constant  and  equal  to  14.7  lb.  per  sq.  in.  For 
instance,  the  weight  of  a  cubic  foot  of  water  at  39°F.  is  62.4 
pounds,  and  the  actual  specific  volume  (volume  of  one  pound)  is 

_L-  =  .016026  cu.  ft.  Hence,  26.79  -f-  .016026  =  1672  cu.  ft. 
62.4 

=  the  occupied  by  steam  formed  from  1  cu.  ft.  of  water  at 

39.2°F.,  the  pressure  of  both  water  and  steam  being  14.7  lb. 

per  sq.  in.  and  the  temperature  of  both  212°F.     This  shows  the 

enormous  expansion  of  water  when  converted  to  steam. 

123.  Any  vapor  in  contact  with  the  liquid  from  which  it  is 
formed  or  having  the  same  temperature  and  pressure  that  it  had 
when  it  was  formed  is  said  to  be  saturated,  if,  in  addition,  it  has 
no  particles  of  the  liquid  entrained  in  it  (mixed  with  it),  it  is  said 
to  be  dry,  and  is  then  spoken  of  as  dry  and  saturated.  If  heat 
be  applied  to  a  vapor  that  is  dry  and  saturated,  its  temperature 
will  increase;  its  volume  will  also  increase,  or  if  not  allowed  to 
increase,  its  pressure  will  increase,  and  the  vapor  is  then  said 
to  be  superheated.  If  superheated  sufficiently,  the  vapor 
will  exhibit  all  the  characteristics  of  a  perfect  gas. 

If  steam  (or  vapor)  that  is  dry  and  saturated  be  subjected  to 
any  additional  pressure,  it  immediately  begins  to  condense;  or, 
if  the  pressure  is  decreased,  it  is  superheated.  In  other  words, 
for  any  particular  pressure,  there  is  only  one  temperature  for 
which  the  steam  will  be  saturated. 

124.  There  is  no  simple  formula  showing  the  relation  between 
the  temperature  and  pressure  of  saturated  steam,  but  the  two 
following  formulas,  in  which  t  =  temperature  and  p  =  pressure, 
will  give  values  accurate  enough  for  all  practical  purposes: 

Between  10  and  250  lb.  per  sq.  in.  abs.  (abs.  means  absolute), 

t  =  241.5  -  ^^  +  1.214p  -  .003385p2  +  .00000448p3     (1) 

For  pressures  between  1  lb.  and  15  lb.  per  sq.  in.,  abs., 

t  =  121.42  -     ^  +  11.892p  -  .5656p2  4-  .0126p3       (2) 

Thus,  for  p  =  10  lb.  per  sq.  in.  abs.,  formula  (1)  gives 
=  241.5  -  ^~  +  1.214  X  10  -  .003385  X  102 

+  .00000448  X  10*  =  193.2° 


84  ELEMENTS  OF  PHYSICS  §4 

For  p  =  250  lb.  per  sq.  in.  abs.,  formula  (1)  gives 
i  =  241.5  —  ^^  +  1.214  X  250  -  .003385  X  2502 

+  .00000448  X  250'  =  401.1° 
The  temperatures  thus  obtained  are  the  ordinary  tempera- 
tures as  recorded  on  a  Fahrenheit  thermometer. 

125.  What  is  called  the  total  heat  of  steam  is  the  number 
of  heat  units  contained  in  one  pound  between  the  temperatures  of 
32°  and  the  temperature  of  the  steam;  it  is  equal  to  the  number 
of  heat  units  in  the  liquid,  called  the  heat  of  the  liquid,  plus 
the  latent  heat  of  vaporization.  Letting  H  =  total  heat,  the 
following  formulas  will  give  the  total  heat  accurately  between 
p  =  14.7  and  p  =  260  lb.  per  sq.  in.  abs.,  and  may  be  used  up 
to  300  lb.  per  sq.  in.  abs.: 

H  =  1175.1  -  ^^  +  .178p  -  .000264p2  (1) 

For  pressures  less  than  14.7  lb.  per  sq.  in.  abs.,  use  the 
following  formula: 

H  =  1119.7  -  ^^  +  3.28p  -  .0754p2  (2) 

For  example,  the  total  heat  in  one  pound  of  dry  and  saturated 
steam  when  the  pressure  is  100  lb.  per  sq.  in  abs.  is,  by 
formula  (1), 

405  0 
B  =  1175.1  -  -j^^  +  .178  X  100  -  .000264  X  1002 

=  1186.2  B.t.u. 
This  result.  1186  B.t.u.  is  the  number  of  heat  units  that  would  be 
given  up  if  1  pound  of  dry  and  saturated  steam  under  a  pressure 
of  100  lb.  per  sq.  in.  abs.  were  condensed  to  water  and  the 
water  were  cooled  to  32°;  it  is  also  the  number  of  heat  units  that 
would  be  required  to  change  1  pound  of  water  at  32°  into  1  pound 
of  dry  and  saturated  steam  at  a  pressure  of  100  lb.  per  sq. 
in.  abs. 

126.  In  Art.  124,  the  temperature  of  saturated  steam  when 
the  pressure  is  10  lb.  per  sq.  in.  was  found  to  be  193. 2°F. ; 
this  is  the  boiling  point  of  water  for  that  temperature.  In  other 
words,  when  the  pressure  is  10  pounds  per  sq.  in.,  the  boiling  point 
is  193.2°  instead  of  212°,  the  boiling  point  when  the  pressure  is 
14.7  pounds  per  sq.  in.  The  boiling  point,  therefore,  depends 
upon  the  pressure,  and  this  explains  why  liquids  boil  at  a  lower 


§4  HEAT  85 

temperature  in  a  partial  vacuum.  It  also  explains  why  a  kettle 
of  water  boils  aways  so  much  faster  when  the  barometer  is  low, 
just  before  a  storm,  than  when  it  is  normal,  when  the  weather  is 
fair.  When  the  pressure  is  100  lb.  per  sq.  in.,  the  boiling 
point  of  water,  calculated  as  above,  is  327.8°. 
Pressure  Boiling  The  little  table  given  in  the  margin  shows 
abs.       point  the  temperature  at  which  water  boils  correspond- 

1  101  8°  mS  to  the  pressure  in  pounds  per  square  inch  given 

2  126.2  in  the  first  column.    For  higher  pressures  up  to  250 

3  141.5  lb.  per  sq.  in.,  the  boiling  points  may  be  found 

4  153.0  by  substituting  the  pressure  in    the   formulas 
}£?•?   (1)  or  (2)  of  Art.  124.     It  is  to  be  noted  that 

o  17U. 1  . 

7  176.9  *ne  temperature  of  saturated  steam  is  the  same 

8  182.9  as  the  boiling  point  of  water  for  the  same  pres- 

9  188.3  sure;  this  follows  from  the  definition  of  saturated 
10         1932  steam. 

127.  In  the  case  of  a  partial  vacuum,  let  i  =  inches  of  vacuum, 
as  shown  by  the  gauge;  then  the  pressure  in  inches  of  mercury 
(the  tension)  is  29.921  —  i.  The  temperature  in  degrees 
Fahrenheit  and  the  pressure  in  pounds  per  square  inch  of  dry 
and  saturated  steam  in  a  partial  vacuum  may  be  calculated  by 
the  following  formulas  up  to  28  inches  of  vacuum,  letting 
m  =  29.921  -  i; 

t  =  121.4  -  —  +  5.84m  -  .1364m2  +  .00149m3  (1) 

m 

H  =  1119.7  -  ^^  +  1.611m  -  .01818m2  (2) 

m 

For  example,  if  the  vacuum  gauge  indicates  22  in.,  the  tem- 
perature   of    the    steam    is,    since    m  =  29.921  —  22  =  7.921, 

t  =  121.4  -  ^||-  +  5.84  X  7.921  -  .1364  X  7.9212 

+  .00149  X  7.9213  =  152°F. 
and  the  total  heat  of  the  steam  is 

H  =  1119.7  -  ^~  +  1.611  X  7.921  -  .01818  X  7.9212 

=  1126.6  B.t.u. 

128.  Melting  and  Boiling  Points  of  Some  Substance. — Every 
known  substance  can  probably  occupy  all  the  three  states  of 
matter;  if  it  is  a  gas,  it  can  be  liquified  and  then  frozen  (solidified)  ; 
if  a  liquid,  it  can  be  frozen  or  vaporized;  if  a  solid,  it  can  be 
liquified  (melted ) and  then  vaporized.     It  is,  of  course,  understood 


86  ELEMENTS  OF  PHYSICS  §4 

that  all  substances  have  not  been  made  to  occupy  all  three  states; 
it  is  only  recently  that  certain  solids  have  been  liquified,  and  not 
all  of  them  have  been  vaporized.     The  following  table  gives  the 

Fusion  Boiling 

point  point 

°F.  CF. 

Alcohol,  absolute  (liquid) —170                             173 

Ammonia  (gas) — 104                           —28 

Aluminum  (solid) 1217 

Carbon  (solid) 6300 

Copper  (solid) 1980                          4190 

Carbon  Dioxide  (gas) —110 

Gold  (solid) 1944 

Helium  (gas) —450 

Hydrogen  (gas) —423 

Iridium  (solid) 4172 

Iron  (solid) 2740 

Lead  (solid) 620                          2777 

Mercury  (liquid) -38                             675 

Nitrogen  (gas) —231 

Osmium  (solid) 3992 

Oxygen  (gas) - 182 

Platinum  (solid) 3191 

Sea  Water  Qiquid) 28 

Silver  (solid) 1760                          3550 

Tantalum 5240 

Tin  (solid) 450                          4118 

Tungsten  (solid) 5430 

Turpentine  (liquid) 14 

Water,  pure  (liquid) 32                             212 

Zinc  (solid) 786                           1684 

fusion  point  or  boiling  point  or  both  of  a  number  of  well-known 
substances,  the  ordinary  state  of  each  substance  being  given  in 
parenthesis.  It  is  to  be  noted  that  the  fusion  point  is  the  tempera- 
ture (Fahrenheit)  at  which  the  substance  changes  from  a  solid  to  a 
liquid  (melts)  or  from  a  liquid  to  a  solid  (freezes),  and  the  boiling 
point  is  the  temperature  at  which  a  liquid  changes  to  a  gas 
(vaporizes)  or  a  gas  changes  to  a  liquid  (liquefies).  In  the  case 
of  carbon,  there  is  apparently  no  liquid  state,  it  changing  directly 
from  the  solid  to  a  gas,  but  the  gas  has  never  been  isolated  and 
examined,  the  temperature  being  too  high. 

129.  Humidity. — There  is  always  a  certain  amount  of  water 
vapor  present  in  atmospheric  air,  and  the  pressure  of  this  vapor 
is  usually  different  from  that  of  the  air.     For  any  particular  tern- 


§4  HEAT  87 

perature  of  the  air,  there  is  a  pressure  for  which  the  water  vapor 
will  be  saturated;  that  is,  any  increase  in  pressure  or  any  decrease 
in  temperature  (however  small)  will  cause  the  vapor  to  condense. 
In  general,  the  water  vapor  is  in  a  superheated  state,  its  pressure 
being  less  than  that  required  for  saturation.  Let  w  be  the  weight 
of  the  vapor  per  cubic  foot  when  the  temperature  of  the  air  is  t, 
and  let  w'  be  the  weight  per  cubic  foot  at  the  same  temperature 
when  the  vapor  is  saturated;  then  w'  will  be  greater  than  w,  and 

w 
the  ratio  —  is  called  the  relative  humidity,  which  has  different 

values    for    different    temperatures    and    pressures. 

Suppose  that  for  some  particular  state  of  the  atmosphere  the 
temperature  is  t  and  the  barometric  pressure  is  6,  and  let  w  be  the 
weight  per  cubic  foot  of  the  water  vapor.  If  the  air  be  cooled, 
both  the  temperature  and  pressure  will  fall;  the  water  vapor  also 
cools  and  after  a  time,  it  will  reach  its  saturation  temperature  t'  and 
saturation  pressure  b' ;  this  temperature  is  called  the  dew  point 
corresponding  to  the  temperature  t  and  pressure  b.  Hot  air  absorbs  a 
greater  amount  of  moisture  than  cold  air,  which  explains  the  heavy 
dews  of  summer.  During  the  night,  the  air  cools  until  the  dew 
point  is  reached,  and  any  further  cooling  causes  the  water  vapor 
to  condense  and  fall  as  dew. 

130.  The  relation  of  humidity  to  health  and  comfort  is  of 
extreme  importance.  Hot,  damp  days  are  oppressive  and  seem 
hotter  than  they  really  are  because  the  relative  humidity  is  so 
high,  the  vapor  is  so  near  the  dew  point,  that  free  evaporation  of 
moisture  from  the  body  is  interfered  with;  on  the  other  hand, 
when  the  relative  humidity  is  low,  the  air  is  dry  and  the  evapora- 
tion of  moisture  from  the  body  is  too  great.  In  houses  and 
offices,  where  the  temperature  is  kept  at  about  68°  or  70°,  the 
best  results  are  obtained  when  the  relative  humidity  is  from 
50%  to  60%;  if  the  relative  humidity  is  less  than  this,  it  will 
require  a  higher  temperature  for  comfort  on  cold  days;  and  if  it  is 
much  less,  the  result  will  be  apparent  by  the  widening  of  cracks 
in  floors  (if  of  wood),  the  loosening  of  the  furniture,  etc.,  and  it 
will  also  have  a  bad  effect  on  the  health. 

In  drying  paper  and  pulp,  the  relative  humidity  in  the  dry 
loft  or  machine  room  is  an  important  factor,  since  the  rate  at 
which  the  air  takes  moisture  from  the  stock  becomes  slower  as 
the  moisture  already  in  the  air  increases;  on  the  other  hand,  the 
amount  of  moisture  in  stored  paper  varies  with  the  moisture 


88  ELEMENTS  OF  PHYSICS  §4 

in  the  air,  and  the  quality  is  likewise  affected  by  this  factor. 
A  well  insulated  roof  and  an  interior  temperature  above  the  dew 
point  of  the  moisture-laden  air  is  required  to  prevent  condensation 
in  the  loft  or  machine  room. 

131.  Foaming  and  Surface  Tension. — If  a  fine  needle  be  placed 
on  the  free  surface  of  a  liquid,  care  being  taken  to  have  its  axis 
parallel  to  the  plane  of  the  free  surface  and  to  lay  it  gently  on  the 
surface,  the  needle  will  float,  and  this  despite  the  fact  that  the 
density  of  the  needle  is  from  to  6  to  8  times  that  of  the  liquid. 
Small,  dust-like  particles  of  anj>-  kind,  though  many  times  as 
dense  as  the  liquid,  will  float  provided  they  come  into  contact 
with  the  liquid  without  the  exertion  of  a  downward  force  that  is 
greater  than  that  due  to  gravit y.  But  if  the  needle  or  the  particles 
are  at  any  time  entirely  submerged,  they  will  sink,  if  of  greater 
density  than  the  liquid.  This  phenomenon  is  due  to  what  is 
called  surface  tension,  which  is  very  closely  related  to  capillarity, 
and  is  explained  as  follows: 

The  molecules  (and  particles)  of  a  liquid  attract  one  another, 
with  the  result  that  they  tend  to  form  compact  masses;  this 
attraction  takes  place  in  all  directions,  and  when  the  masses  are 
small,  they  tend  to  form  little  spheres,  thus  reducing  the  free 
surface  to  a  minimum,  a  sphere  having  the  smallest  surface  for  a 
given  volume.  Thus,  if  a  match  be  dipped  in  a  liquid  and  then 
held  up  so  the  liquid  can  run  down  to  one  end,  a  round  drop  will 
form,  and  if  the  drop  falls,  it  has  the  shape  of  a  perfect  sphere. 
The  size  of  the  drop  will  depend  upon  the  cross-sectional  area  of  the 
end  of  the  match  or  stick  and  upon  the  viscosity  of  the  liquid. 
Viscosity  may  be  defined  as  the  state  of  fluidity;  it  may  also  be 
thought  of  as  internal  friction.  For  instance,  if  a  thin,  flat 
plate  be  drawn  through  a  liquid  in  a  direction  parallel  to  the  plane 
of  its  flat  surface,  the  force  required  to  pull  the  plate  is  a  measure 
of  the  viscosity  of  the  liquid;  it  will  evidently  require  a  greater 
force  to  pull  the  plate  through  molasses  than  through  water; 
hence,  molasses  is  more  viscous — has  greater  viscosity — than 
water. 

Now  when  the  upper  surface  of  a  liquid  is  free,  there  is  no 
attraction  from  above,  but  the  molecules  in  the  immediate  neigh- 
borhood below  the  surface  of  the  liquid  tend  to  pull  downward. 
There  are  also  other  forces  pulling  the  surface  molecules  and 
tending  to  bring  them  as  close  together  as  possible;  that  is,  so  that 
the  area  of  the  exposed  surface  will  be  as  small  as  possible.     This 


§4  HEAT  89 

causes  the  free  surface  to  behave  as  though  an  elastic  skin  were 
drawn  tightly  over  it,  and  it  creates  the  surface  tension  mentioned 
above. 

132.  When  a  liquid  is  violently  agitated,  as  by  rapid  stirring,  by 
the  use  of  an  egg  beater,  by  heating,  etc.,  bubbles  form  on  the  free 
surface;  this  phenomenon  is  called  foaming.  The  bubbles  are  due 
to  surface  tension,  which  enables  them  to  keep  their  form.  The 
amount  and  quality  of  the  foam  that  forms  on  any  liquid  depends 
upon  its  composition,  viscosity,  and  temperature.  The  presence 
of  impurities  will  freqently  cause  the  liquid  to  foam  more  readily 
than  when  pure.  When  pure  water  is  boiled,  bubbles  form  on  the 
free  surface,  but  they  do  not  last,  collapsing  almost  as  soon  as 
they  form;  the  same  thing  happens  when  water  is  violently 
agitated  with  an  egg  beater.  If,  however,  the  water  is  not  pure, 
but  contains  other  substances  in  solution,  making  it  more  viscous, 
the  bubbles  will  be  larger  and  will  retain  their  form  longer;  in 
fact,  they  may  retain  their  form  until  they  collapse  from  eva- 
poration as  in  the  case  of  soap  suds,  etc.  Such  bubbles  are  com- 
monly called  froth. 

133.  In  paper  mills,  the  froth  on  top  of  a  liquid  is  sometimes 
but  the  dried  mechanical  structure  of  evaporated  films,  which, 
upon  examination,  will  be  found  to  be  superimposed  layers  of 
more  or  less  dried  clay,  rosin,  calcium  resinate,  fibers  and  other 
solids,  all  of  which  act  to  form  a  structure  for  fresh  bubbles  to 
adhere  to  and  prevent  the  bubbles  proper  from  breaking.  If  this 
structure  is  allowed  to  accumulate,  it  may  sink,  mix  with  the  pulp 
mixture  or  the  paper  stock,  and  show  up  as  dirty  spots  in  the 
product.  A  spray  of  water  is  sometimes  effective  in  breaking  up 
this  structure,  but  it  is  advisable  in  most  cases  to  skim  it  off  and 
then  endeavor  to  prevent  its  formation  by  one  or  more  of  the 
following  methods: 

(a)  By  spraying  the  surface  (if  it  has  a  tendency  to  foam)  with 
a  fine  needle  spray  of  pure  or  clean  water,  thus  breaking  the 
bubbles  before  they  attain  any  great  size. 

(b)  By  adding  some  solution  to  the  liquid  that  will  increase  its 
surface  tension,  thus  preventing  it  from  breaking  up  into  air 
bubbles. 

(c)  By    adding    clean    water    to    the    mixture. 

(d)  By  lowering  the  temperature  of  the  liquid  and  increasing 
the  temperature  of  the  air  in  the  neighborhood  of  the  liquid. 


90  ELEMENTS  OF  PHYSICS  §4 

(e)  By  mechanical  removal  of  the  froth  as  soon  as  formed  by 
the  use  of  a  skimming  device. 

In  actual  practice,  it  has  been  found  in  some  mills  that  when 
the  temperature  of  the  water  used  on  the  screens  is  lower  than 
25°C.  (77°  F.),  there  is  less  foaming  than  when  the  water  is  above 
this    temperature. 

134.  In  the  case  of  steam  boilers,  what  is  called  priming  occurs 
when  water  bubbles  are  carried  over  with  the  steam.  Consider- 
able water  may  thus  be  added  to  the  steam,  thus  making  it  no 
longer  dry,  and  serious  results  may  follow.  A  relatively  large 
amount  of  water  may  thus  accumulate  in  the  cylinder  of  the  engine 
or  turbine,  and  the  swiftly  moving  piston  or  blades  may  strike 
this,  causing  the  cylinder  to  burst,  since  water  is  practically 
incompressible.  Priming  is  usually  due  to  impurities,  which  may 
be  contained  in  the  feed  water  or  may  be  added  to  it  by  too 
liberal  use  of  boiler  compounds,  added  to  prevent  incrustation. 
When  feed  water  that  is  known  to  foam  must  be  used,  an  analysis 
should  be  made  of  the  water  to  determine  what  impurities  are 
present;  it  is  then  frequently  possible  to  add  some  chemical  that 
will  neutralize  these  impurities  before  the  water  is  fed  into  the 
boilers.  Priming  may  sometimes  be  overcome  by  changing  the 
water  frequently  in  the  boilers,  by  blowing  them  down  and  filling 
them  up  with  fresh  water,  or  by  adding  some  water  glass  (sodium 
silicate),  which  increases  the  surface  tension. 


LIGHT 


RADIANT  ENERGY 

135.  Nature  of  Light. — An  explanation  of  the  dynamical 
theory  of  heat  was  given  in  Art.  117,  and  it  was  there  stated  that 
heat  was  propagated  by  reason  of  vibratory  motions  set  up  in  the 
ether,  which  pervades  all  space  and  all  bodies.  This  vibratory 
motion  results  in  a  series  of  waves,  somewhat  like  those  which  are 
formed  when  a  stone  is  thrown  into  a  pond  when  the  water  is  still. 
The  particles  of  water  move  up  and  down,  communicate  their 
movement  to  the  adjacent  particles,  and  the  wave  moves  outward 
from  where  the  stone  entered  the  water.  Note  that  it  is  the 
particles  of  water  that  move  up  and  down  and  that  it  is  the  wave 
form  that  moves  outward;  the  water  particles  all  return  to  the 


§4  LIGHT  91 

place  from  which  they  started.  This  is  practically  what  takes 
place  in  the  ether,  the  ether  particles  moving  transversely  to 
(across)  the  direction  of  the  wave  movement.  If  a  plane  section 
be  taken  through  a  wave  of  this  kind,  the  result  will  be  something 
like  the  curve  in  Fig.  24,  in  which  the  parts  AaB  =  BbC  =  CcD 
=  etc.  and  AaB  is  symmetrical  to  BbC  with  respect  to  the  point 
B,  BbC  is  symmetrical  to  CcD 
with  respect  to  the  point  C,  p         l 

etc.      The  distance  between  a /b 

any  two  corresponding  points 

on  two  symmetrical  parts  of        p  ; 

the  curve,  as  AC  =  ac  =  bd  FlG  24. 

=  etc.  is  called  a  wave  length, 

here  designated  by  I.     The  highest  points,  b,  d,  etc.,  are  called 

crests,  and  the  lowest  points  c,  e,  etc.,  are  called  troughs.     The 

perpendicular    distance    between    a    crest    and    a   trough,   here 

designated  by  h,  is  called  the  amplitude  of  the  wave. 

136.  All  radiant  energy  is  propagated  (transmitted)  in  the 
form  of  waves.  Heat  and  light  are  both  forms  of  radiant  energy, 
the  sole  difference  between  them  being  in  the  length  of  the  waves; 
if  the  waves  are  too  long,  they  produce  only  the  sensation  of  heat, 
and  if  they  are  too  short,  produce  neither  the  sensation  of  heat 
nor  light,  but  will  produce  chemical  effects  and  are  then  called 
actinic  rays.     Actinic  rays  will  affect  the  photographic  plate. 

137.  Bodies  that  act  as  a  source  of  light,  that  is,  which  are  able 
to  produce  vibrations  in  the  ether  of  such  a  nature  as  to  make 
thern  visible,  are  called  luminous  bodies;  as  a  piece  of  red  hot 
iron,  a  candle  flame,  an  incandescent  electric  lamp,  etc.  Such 
bodies  are  said  to  emit  light.  Some  luminous  bodies  can  be  seen 
to  emit  light  only  in  the  dark  or  in  semi-darkness;  this  is  because 
the  light  of  day  is  so  much  stronger  that  it  overpowers  the  light 
from  the  luminous  body;  thus,  the  light  of  a  lightning  bug,  of  a 
piece  of  wet  phosphorous,  the  light  of  the  stars,  etc.  cannot  be 
perceived  in  full  daylight.  Bodies  that  do  not  emit  light  and 
can  be  seen  only  by  light  that  is  reflected  from  a  luminous  body 
are  called  non-luminous.  Most  bodies  are  non-luminous  under 
ordinary  conditions. 

A  ray  of  light  is  the  line  along  which  light  is  propagated;  rays 
issue  from  the  source  of  light  in  all  directions,  and  if  the  medium 
through  which  they  pass  is  uniform  in  its  structure,  every  ray  of 


92  ELEMENTS  OF  PHYSICS  §4 

light  is  a  straight  line.  A  collection  of  parallel  rays  make  up  a 
beam  of  light;  but  if  the  rays  are  not  parallel,  either  converging 
toward  or  diverging  from  a  point,  they  form  a  cone  or  pencil. 

When  light  passes  through  any  substance  whatever,  a  solid,  a 
liquid,  or  a  gas,  the  substance  is  called  a  medium.  If  light  rays 
pass  freely  through  the  medium,  so  that  objects  are  clearly  seen 
when  looking  through  it,  the  medium  is  transparent;  clear  water, 
air,  ordinary  window  glass,  glassine  paper,  certain  crystals,  etc. 
are  transparent.  If  objects  are  only  faintly  seen  through  the 
medium,  it  is  said  to  be  semi-transparent;  smoked  glass,  a 
foggy  atmosphere,  slightly  muddy  water,  etc.  are  semi-transpar- 
ent. If  light  can  be  seen  through  the  substance,  but  the  form 
of  objects  cannot  be  distinguished,  the  medium  is  called  trans- 
lucent; ground  glass,  colored  glass,  certain  crystals,  etc.  are 
translucent.  Substances  which  allow  no  light  to  pass  through 
them  are  called  opaque;  a  piece  of  iron,  a  stone,  a  stick  of  wood, 
heavy  cardboard,  etc.  are  opaque  mediums.  However,  if 
any  substance  be  cut  in  sufficiently  thin  slices  or  rolled  suffi- 
ciently thin,  it  will  be  more  or  less  translucent.  Opacity  is  a 
desirable  quality  in  printing  papers. 

When  the  medium  is  transparent,  all  the  light  that  strikes  it 
goes  through  it;  but  when  the  medium  is  translucent,  some  of  the 
rays  are  absorbed  and  the  remainder  go  through,  those  that  are 
absorbed  being  transformed  into  heat.  Thus,  a  clean,  plate- 
glass  window  exposed  to  the  sun  in  the  summer  will  be  cool;  but 
if  it  be  covered  with  paint  or  varnish  or  the  surface  be  roughened, 
some  of  the  light  rays  will  be  absorbed  and  the  glass  will  become 
warm — it  may  even  become  hot.  In  the  case  of  opaque  sub- 
stances, all  the  light  rays  striking  it  are  transformed  into  heat. 

138.  Light  Rays  are  Right  Lines. — It  was  stated  in  the  last 
article  that  every  ray  of  light  is  a  right  line,  provided  the  medium 
through  which  it  passes  is  uniform;  thus,  a  ray  of  light  going 
through  a  clear  atmosphere,  clear  water,  glass,  etc.  will  be  straight; 
but,  as  will  be  explained  later,  it  will  not  be  straight  throughout 
its  length,  if  it  passes  through  two  or  more  different  mediums. 
That  the  rays  are  right  lines  is  easily  proved  by  the  fact  that  a 
person  can  see  through  a  straight  tube,  but  cannot  see  through 
it  if  it  is  bent.  So  long  as  the  medium  is  uniform  the  rays  are 
straight ;  but  if  a  ray  passes  through  several  mediums,  it  will  be  a 
broken  line,  that  part  of  the  ray  through  each  medium  being 
straight.     A  ray  can  never  be  a  curved  line. 


§4 


LIGHT 


93 


139.  Velocity  of  Light. — The  speed  of  light  is  so  great  that  it  is 
difficult  to  measure  it  with  any  great  degree  of  exactness.  How- 
ever, the  velocity  has  been  determined  in  a  number  of  different 
ways,  the  most  probable  value  being  186,400  miles  per  second. 
This  is  equivalent  to  300,000  kilometers  =  3  X  1010  centimeters 
per  second,  the  latter  value  being  very  easy  to  remember.  An 
object  moving  at  this  speed  would  be  able  to  go  around  the 
earth  about  7^  times  in  one  second.  The  distance  from  the 
earth  to  the  sun  is  about  93,000,000  miles;  consequently,  light 
received  from  the  sun  requires  93000000  -3-  186400  =  499  seconds 
=  8  min.  19  sec.  to  reach  the  earth;  in  other  words,  the  sun  has 
risen  above  the  horizon  8  min.  19  sec.  before  it  is  seen,  and  it  is 
seen  for  8  min.  19  sec.  after  it  has  set.  Anything  that  happens  on 
the  sun  cannot  be  seen  until  8  min.  19  sec.  after  it  has  occurred. 
The  fixed  stars  are  so  far  away  that  it  takes  over  3  years  to  reach 
the  earth  from  the  nearest  one,  and  some  are  so  distant  that  it 
takes  light  thousands  of  years  to  reach  the  earth. 


SHADOWS 

140.  Shadows  Cast  by  a  Luminous  Point. — If  a  luminous 
body  be  very  small,  so  that  for  practical  purposes  it  may  be 
considered   to  be  a  point,  it  is  called  a  radiant.     Hence,  any 


Fig.  25. 


luminous  body  may  be  considered  as  a  collection  of  radiants 
distributed  over  its  surface.  In  Fig.  25,  let  P  be  a  radiant  and  S 
a  flat,  opaque,  white  screen  having  a  smooth  dull  surface  that  will 
not  reflect  light.  Let  AB  be  a  flat,  opaque  plate  of  irregular 
outline,  situated  between  P  and  S.  Rays  of  light  from  P  shoot 
out  in  straight  lines   in   all   directions — upwards,   downwards, 


94 


ELEMENTS  OF  PHYSICS 


§4 


sideways,  forwards,  and  backwards — and  illuminate  the  screen  S. 
The  rays,  however,  cannot  pass  through  the  plate  AB,  and  the 
result  is  that  there  is  a  dark  spot  CD  on  the  screen,  which  is  called 
the  shadow  of  AB.  The  outline  of  the  shadow  is  of  the  same  form 
as  that  of  the  object  AB;  it  is  really  a  projection  of  AB  on  the 
screen  formed  by  drawing  right  lines  from  the  point  P,  touching 
the  perimeter  of  AB  and  intersecting  the  surface  of  the  screen. 
These  lines  form  a  pencil  (Art.  137)  of  rays,  and  any  ray  included 
within  the  bounding  surface  of  this  pencil  is  stopped  by  the  ob- 
ject and  does  not  strike  the  screen.  This  is  another  proof  of  the 
fact  that  the  rays  of  light  are  straight.  The  outline  of  a  shadow 
formed  by  light  proceeding  from  a  single  radiant  is  sharp  and 
distinct. 

141.  Shadows  Cast  by  a  Luminous  Body. — If  the  shadow  is 
cast  by  a  luminous  body  instead  of  a  single  radiant,  a  somewhat 
different  result  is  obtained.     Referring  to  Fig.  26,  suppose  XY  to 


Fig.  26. 


be  a  luminous  sphere,  RM  an  opaque  sphere,  and  let  *SA~  be  the 
edge  of  a  screen;  then,  if  XR  and  YM  are  tangents  to  the  two 
spheres,  all  radiants  on  the  surface  XAY  emit  rays  of  light  that 
touch  the  object  RM;  but  all  other  radiants  on  the  sphere  A'}" 
emit  raj's  that  do  not  touch  RM.  Considering  the  radiant  X, 
it  forms  a  pencil  RXM.  which  results  in  the  shadow  WD  on 
the  screen;  the  radiant  Y  forms  the  pencil  MYR,  which  results 
in  the  shadow  YC;  the  total  shadow  cast  by  these  two  radiants 
is  CD.  But.  the  part  WC  is  illuminated  with  more  or  less  inten- 
sity by  radiants  situated  between  X  and  }',  the  shadow  becoming 
lighter  and  lighter  as  the  distance  from  IV  toward  C  increases, 
since  rays  from  a  greater  number  of  radiants  strike  the  part  WC 
of  the  screen  above  the  line  ER  than  below  it,  radiants  situated 
between  E  and  Y  also  striking  this  part  of  the  screen.     The  same 


§4  LIGHT  95 

thing  occurs  in  connection  with  the  part  VD  of  the  shadow,  and 
as  a  consequence,  the  edges  C  and  D  are  very  faint  and  indistinct. 
The  part  between  W  and  V  is  not  illumined  by  rays  from  any 
radiant;  this  part  is  equally  dark  throughout  and  is  called  the 
umbra.  The  part  outside  of  the  umbra,  represented  by  WC  and 
VD  graduates  from  the  color  of  the  umbra  to  full  illumination, 
and  is  called  the  penumbra.  The  case  just  described  fulfills  all 
the  conditions  of  a  total  eclipse  of  the  sun.  Here  XY  is  the  sun, 
RM  is  the  moon,  and  the  screen  SN  is  the  earth.  Persons  in  the 
umbra  witness  a  total  eclipse;  those  in  the  penumbra,  see  only 
a  partial  eclipse. 

142.  Brightness  and  Intensity. — The  brightness  of  a  light  does 
not  depend  upon  the  size  or  shape  of  the  luminous  body,  just 
as  the  density  of  a  substance  does  not  depend  upon  the  size  or 
shape  of  the  body.  A  pin  point  of  light  may  be  just  as  bright  as  a 
powerful  search  light.  The  brighter  the  light  the  darker  will 
be  the  shadow  it  casts,  since  the  illuminated  part  around  the 
shadow  will  then  be  brighter  and  there  will  be  a  greater  contrast 
between  the  shadow  and  the  illuminated  portion  of  the  screen. 
The  intensity  of  a  light  however,  depends  upon  the  brightness 
of  the  radiants,  their  number  (and,  consequently,  the  area  of  the 
surface  emitting  the  rays),  and  the  distance  of  the  object  illu- 
minated from  the  source  of  light.  In  fact,  the  intensity  of  light, 
like  the  intensity  of  heat  or  any  other  form  of  radiant  energy, 
varies  inversely  as  the  square  of  the  distance.  The  proof  of  this 
is  exactly  the  same  as  was  given  in  the  case  of  heat,  Art.  118. 

143.  Candlepower. — The  common  standard  for  the  intensity 
of  light  is  one  candlepower,  which  is  the  amount  of  light  emitted 
by  a  sperm  candle  %  inch  in  diameter  when  burning  at  the  rate 
of  120  grains  per  hour.  This  is  called  a  standard  candle.  If, 
therefore,  a  standard  candle  illuminates  a  certain  object  3  ft. 
distant  with  the  same  intensity  as  another  source  of  light  situated 
12  feet  distant,  then,  the  candlepower  of  the  standard  candle  being 
1,  let  c  be  the  candlepower  of  the  other  light.  According  to  the 
the  law  of  the  last  article,  c  may  be  found  from  the  proportion 

144 

c  :  1  =  122  :  32,  or  c  =     Q    =  16  candlepower. 

It  is  to  be  remembered  that  the  proportion  is  an  inverse  one. 

144.  Photometers. — A  photometer  is  an  apparatus  for  meas- 
uring the  relative  intensities  of  light.     Photometers  are  made  in 


96  ELEMENTS  OF  PHYSICS  §4 

various  forms,  two  of  which  will  be  described  here,  as  they  make 
use  of  two  different  principles. 

(a)  Rumford's  Method. — Referring  to  Fig.  27,  T  is  a  horizon- 
tal, flat  surface  on  which  is  erected  a  vertical,  flat,  white  screen  S. 
R  is  an  opaque  rod,  C  is  a  standard  candle  or  other  source  of 
light  whose  intensity  is  known,  and  L  is  a  light  whose  intensity 
or  relative  intensity  as  compared  with  C  is  to  be  measured,  say 
an  incandescent  electric  lamp.  C  casts  a  shadow  of  the  stick  or 
rod  R  at  a,  which  is  illuminated  by  the  light  from  L,  and  L  casts 


Fig.  27. 

a  shadow  of  the  rod  R  at  b,  which  is  illuminated  by  the  light  from 
the  candle  C.  By  moving  C  back  and  forth  along  the  line  Ca, 
it  will  be  found  that  there  is  one  point  for  which  the  two  shadows 
will  be  of  the  same  distinctness;  suppose  this  to  be  the  point  Cin 
the  figure.  Measure  the  distances  Cb  =  m  and  La  =  n;  then,  if 
the  intensity  of  the  candle  be  represented  by  C  and  of  the  lamp 
by   L, 


n 
L  :  C  =  n2  :  m2,  or  L  =  — =  C 

n2 
If  C  is  1  candle  power,  then  L  =  — 5  candlepower.     In  other 

*  m2 

words,  when  two  sources  of  light  illuminate  an  object  with  the  same 
intensity,  these  candle-powers  are  directly  proportional  to  the  squares 
of  their  distances  from  the  object. 

(6)  Bunsen's  Method. — The  essential  part  of  a  Bunsen 
photometer  is  a  disc  of  unglazed  paper  having  a  round  grease  spot 
in  the  center,  thus  making  the  paper  translucent  at  this  spot. 
The  disc  B,  Fig.  28,  is  held  in  an  adjustable  holder,  which  can  be 
moved  along  a  graduated  scale  *S.  A  standard  candle  C  or  other 
source  of  illumination  whose  intensity  is  known  is  placed  on  one 
side  of  the  disc,  and  an  electric  lamp  L  or  other  light  whose  inten- 
sity it  is  desired  to  find  is  placed  on  the  other  side.     If  the  lamp 


LIGHT 


97 


be  turned  out  and  the  candle  be  left  burning,  the  side  of  the  grease 
spot  nearest  the  candle  will  be  lighter  than  the  other  side.  If, 
therefore,  both  lights  are  going,  that  side  of  the  grease  spot  will 
be  the  darker  which  is  turned  toward  the  light  of  less  intensity. 
By  moving  the  holder  B  along  the  scale,  a  point  will  be  found  at 


1 1 1 1 TTi  1 1 1 r f i  it 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ii  1 1  n 1 1 1 1 1 it 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1  p  1 1 1 1 1 1)  n  1 1 1 1 


lO         20         30         40         60         60  70         80         90 


Fig.  28. 


which  both  sides  of  the  spot  are  illuminated  equally.  Then 
measuring  the  distances  m  and  n  between  the  disc  and  C  and  L, 
respectively, 


L=  -2C 


the  same  result  as  was  obtained  by  Rumfords'  method. 


REFLECTION  OF  LIGHT 

145.  Reflection  from  Plane  Surfaces. — Let  P,  Fig.  29,  be  a 
polished  plane  surface,  as  a  mirror,  polished  steel,  etc.,  and  let  C 
be  a  radiant.  Suppose  a  ray  of  light  from  C  to  strike  the  surface 
P  at  A ;  instead  of  being  absorbed,  the  light  will  be  reflected  at 
A  along  the  line  AE,  and  if  the  eye  is  situated  at  G,  the  reflected 
ray  will  enter  it.  Pass  a  plane  through  A  and  C  perpendicular 
to  plane  P;  it  will  be  found  that  the  reflected  ray  AE  lies  in  this 
plane,  and  a  perpendicular  P'A  at  A  also  lies  in  this  plane.  The 
angle  CAP'  is  called  the  angle  of  incidence,  and  the  angle  P' AE 
is  called  the  angle  of  reflection.  It  has  been  proved  by  repeated 
experiments  and  measurements  that  the  angle  of  incidence  is 
always  equal  to  the  angle  of  reflection,  that  is,  CAP'  =  P'AE  and 
P'A  bisects  the  angle  CAE.  Hence,  to  draw  a  reflected  ray 
from  a  plane  surface,  draw  a  perpendicular  to  the  surface  at  the 

7 


98 


ELEMENTS  OF  PHYSICS 


§4 


point  where  the  incident  ray  strikes  it,  the  point  A  in  Fig.  29; 
then  from  A  draw  AE  so  that  angle  P'AE  =  P'AC,  and  AE  will 
be  the  direction  of  the  reflected  ray.  For  any  other  ray,  as  CB, 
draw  a  perpendicular  at  B,  and  then  draw  BF  so  that  the  angle 
included  between  BF  and  the  perpendicular  BP"  will  equal  the 
angle  included  between  BC  and  the  perpendicular. 


Fig.  29. 

The  line  XY  passing  through  the  point  A  is  the  trace  of  the 
plane  that  includes  AC  and  AE,  and  if  CD  be  drawn  perpendicular 
to  XY  it  will  lie  in  this  perpendicular  plane,  since  it  will  then  be 
parallel  to  P'A.  Produce  EA  until  it  intersects  CD  in  D;  then, 
from  geometry,  angle  YAE  =  angle  XAD,  and  since  YAE 
=  XAC,  XAD  =  XAC,  the  triangle  CAD  is  isosecles,  and  CA 
=  DA.  With  the  eye  in  the  position  shown,  the  point  C  will 
appear  to  be  located  at  D;  that  is,  the  apparent  position  of  the 
radiant,  as  viewed  by  the  eye,  will  appear  to  be  as  far  below  the 
plane  P  as  it  actually  lies  above  the  plane  P;  and  this  will  be  the 
case  no  matter  where  the  eye  is  situated  above  the  plane  of  re- 
flection, since  the  triangle  formed  by  CD  and  the  sides  drawn  from 
C  and  D  will  always  be  isosceles  and  the  length  of  the  base  CD  is 
not  changed.  This  will  still  be  true,  even  when  the  eye  is  di- 
rectly over  C  and  the  triangle  has  become  a  right  line  of  which 


§4  LIGHT  99 

CD  is  part.  This  effect  will  be  apparent  in  the  reflections  pro- 
duced in  any  plane  mirror;  the  object  reflected  will  always  appear 
as  far  back  of  the  mirror  as  it  actually  is  in  front  of  it. 
146.  Reflection  from  Curved  Surfaces. — The  law  of  reflection 
for  curved  surfaces  is  the  same  as  for  plane  surfaces,  provided 
the  angle  of  incidence  and  reflection  are  those  included  between 
the  incident  and  reflected  rays  and  the  normal  to  the  surface  at 
the  point  of  their  intersection.  A  normal  to  a  surface  at  any 
point  is  a  perpendicular  drawn  to  a  tangent  to  the  surface  at  that 


jz*c* 


point.  For  instance,  a  radius  is  always  normal  to  a  circle  or 
sphere  at  the  point  where  it  intersects  the  circle  or  sphere, 
because  a  tangent  drawn  at  that  point  is  perpendicular  to  the 
radius.  For  other  curved  surfaces,  special  geometrical  construc- 
tions are  required  to  draw  a  tangent  at  any  particular  point;  but 
when  a  tangent  has  once  been  drawn,  a  perpendicular  to  the  tan- 
gent at  the  point  of  tangency  will  be  normal  at  that  point. 
Having  drawn  a  normal  to  the  surface  at  the  point  where  the  inci- 
dent ray  strikes  it,  measure  the  angle  between  the  incident  ray 
and  the  normal,  pass  a  plane  through  the  lines  defining  the  inci- 
dent ray  and  the  normal,  and  lay  off  in  this  plane  on  the  other  side 
of  the  normal  an  angle  equal  to  the  one  measured;  the  side  of  this 
angle  will  have  the  same  direction  as  the  reflected  ray  and  will  coin- 


100 


ELEMENTS  OF  PHYSICS 


§4 


cide  with  it.  Thus,  suppose  MN,  Fig.  30,  to  be  a  section  of  a 
spherical  surface  whose  center  is  0;  let  C  be  a  radiant  and  C'A'  a 
ray.  Draw  OA'  and  it  will  be  a  normal  to  the  surface  at  A';  draw 
A'B'  so  it  will  make  an  angle  B'A'O  equal  to  the  angle  C'A'O, 
and  A'B'  wiU  be  the  reflected  ray  when  C'A',  OA',  and  B'A'  all 
lie  in  the  same  plane.  Again,  if  C"  be  a  radiant  and  C'A"  a 
ray,  draw  OA'"  through  0  and  A";  it  will  be  a  normal  to  the 
surface  at  the  point  A".  Lay  off  A"B"  so  angle  A'"A"B" 
=  A'"A"C",  and  A"B"  will  be  the  reflected  ray.  The  arrow- 
heads indicate  the  direction  of  the  incident  and  reflected  rays. 
If  that  part  of  the  curved  surfaces  facing  toward  A'"  be  pol- 
ished, thus  reflecting  all  the  rajrs  striking  it,  and  the  eye  be 
placed  anywhere  along  A"B",  the  radiant  C"  will  appear  as 
though  located  at  D.  To  find  the  point  D,  draw  TT'  tangent 
to  the  surface  at  A"  and  in  the  same  plane  as  A"B"  and  A"C"; 
draw  B"C"  parallel  to  TT'  and  CD  parallel  to  OA'";  pro- 
duce B"A"  until  it  intersects  C"D  in  D,  which  is  the  required 
point. 

147.  Diffusion  of  Light. — Suppose  a  beam  of  light  to  strike  a 
smooth,  polished  plane  surface,  say  a  plane  mirror,  as  shown  at 
(a),  Fig.  31.  The  rays  will  all  be  reflected  in  parallel  lines  and 
will  illuminate  any  object  that  they  strike  only  to  the  extent 
of  an  area  equal  to  the  area  of  a  cross-section  of  the  beam. 


(a) 


Fio.  31. 


Now  suppose  the  reflecting  surface  to  be  comparatively  rough, 
like  ground  glass,  unglazed  white  paper,  etc.  The  incident  rays 
will  all  be  parallel  as  before,  but  the  reflected  rays  will  take 
practically  every  direction,  as  indicated  in  (b),  with  the  result 
that  every  part  of  the  object  will  be  illuminated  and,  in  addition, 
all  parts  of  the  room  as  well.  The  light  is  then  said  to  be  diffused ; 
it  will  illuminate  no  part  of  the  room  with  the  same  intensity  as  in 
(a),  but  it  will  be  softer,  less  trying  on  the  eyes,  and  will  illuminate 
a  far  greater  area.  The  manner  in  which  book  paper  reflects 
light  is  an  important  consideration.  Very  smooth,  glazed  paper 
is  trying  to  the  eyes,  especially  under  artificial  light. 


§4  LIGHT  101 

148.  Visibility  of  Objects. — In  the  case  of  a  self-luminous 
body,  the  light  rays  are  transmitted  directly  to  the  eye,  and  the 
form  of  the  body  is  sharply  outlined.  When  a  body  is  non- 
luminous,  however,  as  is  usually  the  case  with  visible  objects, 
it  can  be  seen  only  by  reflected  light,  which  is  almost  invariably 
diffused,  the  result  being  that  a  non-luminous  body  is  not  usually 
as  distinct  as  a  luminous  one.  The  diffusing  surface  may  be 
considered  as  made  up  of  an  extremely  large  number  of  elementary 
areas,  each  of  which  reflects  light  in  all  directions  from  a  lumi- 
nous source,  thus  making  visible  the  outline  of  the  body,  no 
matter  where  the  eye  is  placed  within  the  limits  of  visibility. 
A  perfectly  reflecting  surface  or  a  transparent  one  is  not  visible 
to  the  eye.  The  water  in  the  gauge  glass  of  a  steam  boiler  cannot 
be  seen  when  clear;  if  the  gauge  glass  is  only  partly  full,  the 
steam  above  it  cannot  be  seen;  and  the  only  reason  that  the 
surface  of  the  water  in  the  glass  can  be  seen  is  because  of  the 
difference  in  refractive  powers  of  water  and  the  air  or  steam  above 
it.  A  polished  reflecting  surface,  like  a  plate-glass  mirror,  is  not 
visible,  since  the  reflected  rays  are  the  same  in  all  respects  as  the 
incident  rays,  and  only  the  luminous  source  is  seen.  This  is 
proved  by  the  fact  that  if  the  wall  of  a  room  were  made  one  laige 
plane  mirror,  a  person  walking  toward  it  will  not  recognize  the 
wall  until  he  strikes  it.  A  black  body  does  not  reflect  rays  to  the 
eye,  but  is  made  visible  by  rays  reflected  from  its  surroundings. 


REFRACTION  OF  LIGHT 

149.  Index  of  Refraction. — In  Fig.  32,  H'H  represents  the 
water  level  of  a  body  of  water,  and  AO  is  a  light  ray  that  strikes 
the  water  (a  plane  surface)  at  0.  A  part  of  the  light  is  reflected 
along  the  line  OA 'and  the  rest  enters  the  water.  Now  instead 
of  continuing  through  the  water  in  the  direction  of  the  incident 
ray  AO  and  along  OB',  the  ray  is  deflected  at  0  and  takes  the 
direction  OB.  The  angle  B'OB  is  called  the  angle  of  deviation, 
and  the  change  in  direction  of  the  incident  ray  when  passing 
through  the  water  is  called  refraction;  AO  is  called  the  incident 
ray  and  OB  the  refracted  ray.  Through  0,  draw  C'C  perpendicu- 
lar to  H'H;  then  AOC  is  the  angle  of  incidence  and  BOC 
is  called  the  angle  of  refraction.  It  is  to  be  noted  that  the  incident 
ray  and  the  refracted  ray  are  on  opposite  sides  of  the  perpendicular 
drawn  at  the  point  of  incidence  0;  also,  the  angle  of  refraction 


102 


ELEMENTS  OF  PHYSICS 


increases  as  the  angle  of  incidence  increases,  and  vice  versa; 
hence,  when  the  incident  ray  is  perpendicular  to  the  surface  H'H, 
there  is  no  refraction,  both  the  incident  ray  and  the  refracted  ray 
being  normal  to  H'H;  but  for  any  other  position,  a  ray  is  always 
refracted  when  it  enters  or  leaves  a  different  medium.  With  but 
few  exceptions,  when  a  ray  passes  from  a  transparent  medium  of 
less  density  (as  air)  through  or  into  one  of  greater  density  (as 
water,  glass,  etc.),  the  refracted  ray  is  inclined  towards  the  normal 
drawn  through  the  point  of  incidence;  thus,  in  Fig.  32,  OB  is 


inclined  towards  OC,  and  BOC  is  less  thanAOC.  Consequently, 
also,  when  a  ray  immerges  from  a  medium  of  greater  density  into 
one  of  less  density,  the  refracted  ray  is  inclined  away  from  the 
normal;  thus,  in  Fig.  32,  if  BO  be  the  incident  ray,  it  will  not 
follow  the  dotted  line  OA"  (which  is  BO  produced),  but  will  take 
the  direction  OA,  being  inclined  farther  from  the  normal  C'C, 
and  AOC  is  greater  than  A"0C  =  BOC. 

At  any  point  m  on  OA,  draw  a  perpendicular  mn  to  OC; 
then  mnO  is  a  right  triangle,  right-angled  at  n,  and  Om  is  the 
hypotenuse.  The  side  mn  is  called  the  side  opposite  the  angle  0. 
Now,  in  any  right  triangle,  the  ratio  of  the  side  opposite  an  angle 
to  the  hypotenuse  is  called  the  sine  of  that  angle;  this  may  be 
expressed  as 

side  opposite 


sine  = 


hypotenuse  ' 
For  example,  in  the  right  triangle  mnO,  sine  0 


mn 
Om 


,  sine  m  = 


On 
Om' 


§4  LIGHT  103 

,    .  Om 

and  sine  n  =  ^  =  1,  that  is,  the  sine  of  90°  (a  right  angle)  is  1. 

Here  sine  0,  sine  m,  etc.  mean  sine  of  angle  0,  sine  of  angle  m,  etc. 
It  from  the  point  p,  a  perpendicular  be  drawn  to  C'C  pqO  will 
be  a  right  triangle,  right-angled  at  q,  and  pO  will  be  the  hypotenuse ; 
then  sine  pOg  =  g|.     It  has  been  found  by  experiment  and 
observation  that  the  ratio  of  the  sines  of  the  angles  of  incidence  to 
the  corresponding  angles  of  refraction  is  constant;  that  is    no 
matter  what  the  inclination  of  the  incident  ray  may  be,  the  ratio 
ot  the  sine  of  the  angle  of  incidence  to  the  sine  of  the  angle  of 
refraction  always  has  the  same  value,  which  is  called  the  index 
of  refraction.     In  Fig.  32, 
sine  AOC  (angle  of  incidence)      mn__pq       mnXOp     mn 
sine  BOC  (angle  of  refraction)      0m'  0p~  p~q~X0m~  =  pj 
when  Op  is  made  equal  to  0m,  as  is  usually  done. 

The  index  of  refraction  varies  for  different  mediums;  in  tables 
that  give  the  values  of  the  index  for  various  mediums,  the  ray 
of  light  is  supposed  to  pass  from  a  vacuum  into  the  given  medium 
and  the  indexes  of  refraction  are  called  absolute  indexes  It 
may  be  remarked  that  the  absolute  index  of  refraction  for  air  is 
so  nearly  1  that  it  may  be  considered  as  1  in  practice;  in  other 
words,  there  is  practically  no  refraction  in  air. 

150.  If  the  ray  pass  from  a  denser  medium  into  a  rarer  one  as 
from  water  into  air,  the  refracted  or  emergent  ray  is  inclined 
away  from  the  normal.  Thus,  referring  to  Fig.  33,  let  AO  be  a 
ray  that  leaves  the  water  at  0;  it  is  refracted  and  has  the  direc- 
tion OB  m  the  air.  If  the  ray  had  the  direction  CO,  the  refracted 
ray  would  have  the  direction  OD,  and  would  lie  in  the  upper 
surface  of  the  water.     The  angle  COP'  for  which  the  emergent 

mu  r.um  SUrfaCG  °f  COntact  between  the  tw°  mediums  is 
called  the  critical  angle;  for  water,  its  value  is  48°  33',  say  48W0 
If  the  incident  ray  make  a  still  greater  angle  of  incidence,  it  will 
not  leave  the  water  at  all,  but  will  be  reflected  back  in  the  same 
manner  as  though  the  surface  of  the  water  were  a  mirror  This 
is  called  total  reflection,  because  no  part  of  the  ray  is  absorbed  by 
the  reflecting  surface;  thus,  the  ray  EO  is  reflected  along  OF 
lhis  phenomenon  applies  to  other  transparent  substances  as 
glass  and  quartz,  and  is  made  use  of  in  the  construction  of  some 
optical  instruments.  A  prism  face  can  thus  be  made  to  reflect 
certain  rays  and  to  transmit  others. 


104 


ELEMENTS  OF  PHYSICS 


§4 


The  direction  of  a  ray  of  light  in  a  medium  of  uniform  density 
is  straight,  that  is,  it  is  a  right  line.  If  a  ray  pass  from  one  me- 
dium, as  air,  through  another,  as  glass,  and  on  into  the  first 
again,  the  ray  will  be  refracted  twice  and  the  incident  and  emer- 
gent rays  will  be  parallel  if  the  surfaces  of  the  refracting  sub- 
stance be  parallel. 

It  is  because  of  refraction,  that  a  straight  stick  or  rod  appears 
bent  when  partly  submerged  in  water.  For  instance,  the  stick 
S,  Fig.  33,  will  appear  as  fgh  instead  of  as  fga  when  the  eye  is 


Fig.  33. 

situated  at  e.  If  the  entire  stick  were  in  air,  a  ray  from  a  to  the 
eye  would  lie  along  ae;  but  on  account  of  the  refraction,  the  ray 
is  bent,  and  takes  the  direction  be,  and  does  not  enter  the  eye. 
The  ray  from  a  that  enters  the  eye  has  the  direction  ad  in  the 
water  and  de  in  the  air;  this  makes  the  end  of  the  stick  appear 
to  be  at  h,  h  being  on  the  line  ed  produced.  As  h  is  nearer  the 
surface  than  a,  the  depth  of  any  object  below  the  surface  does 
not  seem  as  great  as  it  really  is  (unless  the  eye  is  in  a  vertical  line 
over  the  object.  A  small  fish  lying  in  the  water  at  a  will  appear 
to  be  at  h  when  the  eye  is  anywhere  along  the  line  ae. 

When  a  ray  of  light  is  capable  of  being  refracted,  it  is  said  to  be 
refrangible,  the  word  refrangible  being  used  instead  of  refract- 
able.  As  will  presently  be  shown,  certain  kinds  of  fight  rays  are 
more  refrangible  than  others. 

151.  Refraction  through  a  Prism. — Let  ABC,  Fig.  34,  be  a 
cross-section  of  a  transparent  colorless  glass  prism,  and  let  M 
be  a  radiant.     Draw  CD  bisecting  the  angle  C;  suppose  the  prism 


§4 


LIGHT 


105 


to  be  so  located  that  CD  is  vertical.  It  will  evidently  be  possible 
to  place  the  prism  in  such  a  position,  with  CD  vertical,  that  a  ray 
MN  will  be  refracted  along  NP,  perpendicular  to  CD;  NP  will 
then  be  horizontal.  When  the  refracted  ray  emerges  from  the 
prism  at  P,  it  will  take  the  direction  PQ,  the  angle  BPQ  being 


Fig.  34. 


equal  to  ANM.  It  is  also  possible  to  place  the  prism  so  that  a 
ray,  when  refracted  will  meet  the  opposite  face  at  such  an  angle 
that  the  ray  will  be  either  absorbed  in,  or  reflected  by,  the  surface 
(Art.  150). 


LENSES 
152.  A  lens  is  a  refractive  medium  bounded  by  two  surfaces 
through  which  light  passes;  one  of  the  surfaces  is  spherical  (that 
is,  a  part  of  a  spherical  surface)  and  the  other  is  either  a  plane  or  a 


SB 


Fig.  35. 


part  of  a  spherical  surface.  Lenses  are  classified  according  to  the 
nature  of  these  surfaces,  there  being  six  different  kinds,  shown  in 
section  in  Fig.  35  and  numbered  from  1  to  6.     Before  naming 


106  ELEMENTS  OF  PHYSICS  .  §4 

them,  refer  to  (7),  which  shows  two  curves  P  and  Q  and  a  right 
line  MN  that  may  be  considered  to  be  the  trace  of  a  plane,  as  the 
surface  of  a  flat  table  top.  The  curve  P  may  be  regarded  as 
the  inside  surface  of  a  saucer;  it  is  said  to  be  concave  with  respect 
to  the  plane  (or  line)  MN;  the  curve  Q  may  be  regarded  as 
the  inside  surface  of  a  saucer  that  has  been  turned  upside  down, 
and  it  is  said  to  be  convex  with  respect  to  the  plane  (or  line) 
MN.  Bearing  these  definitions  in  mind,  lens  (1)  is  a  double 
convex  lens,  because,  if  laid  on  a  flat  surface  with  side  B  down, 
the  upper  side  A  will  be  convex;  and  if  laid  with  side  A  down, 
the  upper  side  B  will  be  convex.  Lens  (2)  is  called  plano-convex, 
because,  if  laid  on  a  plane  surface  with  side  B  down,  the  upper 
surface  (which  is  flat)  will  be  a  plane,  and  if  laid  with  side  A 
down,  the  upper  surface  B  will  be  convex.  Lens  (3)  is  crescent- 
shaped  and  is  called  a  meniscus;  if  laid  with  side  B  down,  the 
upper  surface  A  will  be  concave,  and  if  turned  upside  down  (rest- 
ing on  the  points  a  and  b),  the  upper  surface  B  will  be  convex. 
Lens  (4)  is  called  double-concave,  because,  whichever  side  is 
down,  the  other  will  be  concave.  Lens  (5)  is  called  plano- 
concave, side  A  being  flat  and  B  concave  to  it.     Lens  (6)  is  called 


Fig.  36. 


concavo-convex,  because,  if  side  B  is  down,  side  A  will  be  con- 
cave, and  if  side  A  is  down,  side  B  will  be  convex.  Strictly 
speaking,  lens  (3)  might  also  be  called  concavo-convex,  but  on 
account  of  its  crescent  shape,  it  is  called  a  meniscus,  and  the  term 
concavo-convex  is  restricted  to  the  form  shown  in  (6). 

Referring  to  Fig.  36,  in  both  (a)  and  (b),  C  and  C  are  the  cen- 
ters of  the  spherical  surfaces,  CA  being  the  radius  of  the  surface 
A  and  C'B  the  radius  of  the  surface  B.  The  line  CC  joining  the 
two  centers  is  called  the  principal  axis  of  the  lens;  in  Fig.  35,  XY 
is  the  principal  axis.  The  two  surfaces  A  and  B  are  called  the 
faces  of  the  lens.  Every  lens  has  a  point  0,  Fig.  36,  (a)  and  (6), 
called  the  optic  center,  which  is  situated  between  the  faces  in  the 


§4  UGHT  107 

Cc7*ldhC<B°»rie  C0DWK0r  d0UWe  C°nCaVe  knSeS-      H  th*  •*■ 

LA  and  C  B  are  equal,  the  optic  eenter  is  midway  between  the 
faces;  m  any  case,  it  lies  on  the  principal  axis 

nrii,snX?  '7  C7idered  f  comP^d  of  an  i„6nite  number  of 
P  isms  the  explanations  in  Art.  151  therefore  cover  the  principle 
of  the  lens.  I„  general,  the  rays  transmitted  by  any  coiZZ 
are  caused  to  converge,  while  those  transmitted  by  a  cZZ  5 
are  caused  to  diverge.  The  effects  of  a  double  convex  len  (wh Z 
s  the  most  common  form)  and  of  a  double  concave  len    on  a 

WFi!  T  r?'!eltt0,the  prfnciPal  axis  ™  s»°™  in  (  )  and 
(b),  Fig.  36.  In  (a),  if  the  lens  is  properly  ground,  the  refracted 
rays  meet  ma  point  F,  which  is  called  the  principal  f  o  us  ™ (1) 
the  emergent  rays  cannot  meet;  but  if  their  lines  of  direction  be 
produced  backwards,  they  will  meet  in  a  point  F,  as  shown  which 

no  nt  ^tTh  rS''  ^emergent  rayS  ""*»  *  come  fronftht 
pomt.  The  distance  FO  between  the  principal  focus  and  the 
optic  center  is  called  the  focal  length.  The  distance  ab  fe  cal  ed 
the  dtameter  of  the  lens.     The  thinner  the  lens  for  the  same 

and  C  B,  and  the  longer  will  be  the  focal  length 

held  mZnfoff  fr°m  fe  ™"  ^  ParaIleL  If  a  c0—  '-s  be 
field  in  fiont  of  a  piece  of  paper  in  such  a  manner  that  the  li«rht 

ays  from  the  sun  are  parallel  to  the  principal  axis  of  the  Ins  and 

reasedaunln  r^  ""  **"■  ~*  ""  ^  "  *«"  "rdl 
Zlocul  It  rt  ^  C°nVe,'ge  '"  a  P0int' the  P°int  is  th"  Princi- 
pal focus  and  the  rays  are  said  to  be  focused.    If  the  paner  and 

ens  are  then  both  kept  stationary,  the  paper  w.ll    e  y  so„n  get 

tak?  rf  7?!nWh7  th°  T  ^  f°CUSed  a»d  '  »*^E  « 
take  fire.     A  lens  acting  ,„  this  manner  is  called  a  burning  glass 

If  a  luminant  be  placed  at  the  principal  focus,  the  rays  from  U 

w.l  emerge  parallel  to  the  principal  axis.     On  the  otChand 

if  the  light  be  placed  beyond  the  principal  focus,  the  emergent 

rays  will  converge  and  form  an  image;  this  can  be  easily  verified 

by  holding  a  lens  between  the  flame  of  a  candle  and  a  shtet  of 

paper,  moving  the  lens  and  the  paper  until  they  are  so  adjusted 

with  respect  to  the  flame  that  a  distinct  image  of  the  flame  can  be 

seen  on  the  paper.    This  is  the  principle  of  the  earner"  in  which 

"in   ecus  '"tT  *°  gCt  tHe  ^^  dfafinCt'  U-  ,0  ■-  ^  obj  ct 
in  iocus       The  eye  is  a  camera;  but,  in  this  case,  instead  of  the 

lens  moving   ,t  automatically  becomes  thinner  or  thicker   ac 

cording  as  the  object  is  far  or  near  from  the  eye 


108  ELEMENTS  OF  PHYSICS  §4 

153.  The  Magnifying  Lens  or  Simple  Microscope. — Referring 
to  Fig.  37,  L  is  a  double  convex  lens  andF  is  its  principal  focus. 
An  object  AB  is  between  the  lens  and  F,  and  the  rays  enter  the 
eye  on  the  other  side  of  the  lens.  The  effect  is  as  though  the 
rays  came  directly  from  a  magnified  image  at  A'B'.  Such  a 
lens  is  a  simple  microscope  or  magnifying  glass.  Note  that  the 
apparent  positions  of  A  and  A'  and  of  B  and  B'  are  determined 
by  extending  the  ray  AO,  which  passes  through  the  optic  center 
0  of  the  lens,  and  tracing  the  ray  BD  which  is  parallel  to  the 


c" 


Fig.  37. 


principal  axis  and  consequently  a  limiting  ray,  until  it  meets  AO 
produced  at  G.  Similarly,  the  limiting  ray  AC  and  the 
undeviated  ray  BO  are  traced  until  they  meet  at  E.  If,  now,  the 
emergent  ray  HE  and  the  ray  AOG  are  produced  backwards, 
they  will  meet  at  A',  and  IG  and  BOE  produced  backwards 
will  meet  at  B\  All  other  points  in  AB  produce  images  between 
A'  and  B',  and  this  complete  image  of  the  points  in  AB  is  what  the 
eye  sees.  It  will  be  noticed  that  if  the  emerging  rays  HE  and  IG 
are  produced,  they  will  meet  at  the  principal  focus  F',  since  AC 
and  BD  are  parallel  to  the  principal  axis.  The  lens  must  be  so 
held  that  the  points  E  and  G  will  fall  on  the  lens  of  the  eye. 
The  compound  microscope  is  an  arrangement  of  several  lenses 
that  greatly  magnifies  the  object.  It  is  used  for  measuring  the 
length  and  thickness  of  fibers,  counting  bacteria  in  milk,  etc. 
The  limit  of  magnification  for  the  simple  microscope  is  about 


§4 


LIGHT 


109 


100  diameters;  the  limit  for  the  compound  microscope  has 
not  as  yet  been  reached,  but  a  power  of  1200  diameters  is  not 
uncommon,  and  some  instruments  have  a  power  of  from  2500  to 
3000  diameters. 


DISPERSION  OF  LIGHT 
154.  The  Solar  Spectrum.— Suppose  that  ab  is  a  narrow  slit  in 
an  opaque  screen  R,  Fig.  38  and  that  a  thin  beam  of  sun  light 
passes  through  the  slit,  impinges  on  the  glass  prism  P  along  cd  is 
refracted  to  ef,  from  whence  it  emerges.  The  light  rays  from'  ab 
to  cd  and  from  cd  to  ef  are  parallel;  but  after  leaving  the  prism 
at  ef,  they  form  a  pencil  of  rays  egjihf  having  the  shape  of  a  wedge 


Fig.  38. 

They  will  intersect  the  screen  S  in  a  rectangle  gjih  and  exhibit  all 
the  colors  of  the  rainbow.  The  colored  outline  thus  formed 
on  the  screen  S  is  called  the  solar  spectrum;  the  colors  are  called 
the  colors  of  the  spectrum  or  the  prismatic  colors. 

166.  An  examination  of  the  spectrum  shows  seven  prominent 
colors,  viz.:  red,  orange,  yellow,  green,  blue,  indigo,  and  violet; 
these  are  called  the  seven  prismatic  colors.  Moreover,  they  are 
always  arranged  in  the  above  order,  with  the  red  nearest  the 
line  ./where  the  rays  emerge  from  the  prism,  and  the  violet 
farthest  away.  The  first  letters  of  the  words  designating  each  of 
the  seven  colors  are  given  on  the  margin  of  the  spectrum  in  Fig 
37,  and  show  the  position  of  colors  in  the  spectrum;  reading 
them  downwards,  it  will  be  noticed  that  they  form  the  word 
vibgyor,  which  will  assist  in  remembering  the  colors  and  their 
order.     It  must  not  be  inferred  that  these  are  the  only  colors  or 


110  ELEMENTS  OF  PHYSICS  §4 

that  they  are  sharply  differentiated;  as  a  matter  of  fact,  the 
spectrum  shows  every  color,  shade,  hue,  and  tint,  all  blending 
into  one  another,  but  the  colors  mentioned  are  the  most  promi- 
nent, and  their  order  is  always  as  here  given. 

The  cut  shows  that  the  rays  entering  the  red  part  of  the  spectrum 
are  not  as  refrangible  as  those  entering  the  regions  above  the  red, 
the  violet  rays  being  much  more  refrangible  than  the  red  rays. 
As  the  result  of  experiment  and  careful  measurement  (by  methods 
that  cannot  be  explained  here),  it  has  been  found  that  the  wave 
lengths  of  the  ra}rs  differ  in  accordance  with  the  color  produced, 
the  wave  length  of  the  violet  rays  being  the  shortest  and  of  the  red 
rays  the  longest.  Since  the  speed,  or  velocity,  of  all  rays  is  the 
same,  it  follows  that  the  longer  the  wave  the  smaller  will  be  the 
number  of  vibrations  per  second  in  the  ether;  and  the  shorter  the 
wave  the  greater  will  be  the  number  of  vibrations  per  second. 
Hence,  the  red  rays  not  only  have  a  greater  wave  length  than  the 
violet  rays,  but  the}'  also  make  a  smaller  number  of  vibrations 
per  second.  Color  is  therefore  due  to  the  wave  length;  and  if 
the  wave  length  is  greater  than  that  of  the  red  ray  or  is  shorter 
than  that  of  the  violet  ray,  the  ray  will  make  no  impression  on  the 
eye.  that  is,  it  cannot  be  seen.  It  is  likewise  evident  that  each 
color  has  its  own  index  of  refraction. 

156.  White  Light  and  the  Primary  Colors. — Since  all  colors 
except  white  and  black  are  included  in  the  spectrum,  it  is  now 
clear  that  white  light  as  received  from  the  sun  is  a  compound  or 
mixture  of  all  the  colors.  Black  is  not  a  color  at  all — it  indicates 
the  absence  of  color.  The  process  of  resolving  light  into  its 
component  colors  is  called  dispersion  or  decomposition  of  light. 

None  of  the  colors  of  the  spectrum  can  be  decomposed  into 
any  other  color  or  combination  of  colors;  this  can  be  shown  by 
cutting  a  slit  anywhere  in  the  screen  covered  by  the  spectrum,  as 
mn  in  the  blue  region,  Fig.  38.  Allowing  the  rays  to  pass  from 
this  slit  through  the  prism  P'  to  the  screen  S,'  both  the  incident 
and  emergent  rays  will  be  parallel,  and  the  color  on  the  screen  S' 
will  be  exactly  the  same  as  that  of  the  slit  on  S.  From  this  it 
will  be  seen  that  each  color  has  its  own  index  of  refraction. 

If  the  screen  S  be  replaced  by  a  series  of  plane  mirrors  so  arranged 
that  they  will  reflect  the  various  rays  received  from  the  prism  P 
to  a  common  point,  called  the  focus,  the  light  at  the  focus  will 
be  white,  thus  proving  that  a  combination  of  all  the  colors  of  the 
spectrum  produces  white. 


§4  LIGHT  111 

157.  Visible  and  Invisible  Rays. — Those  rays  which  produce 
the  sensation  of  color,  those  whose  wave  lengths  are  not  longer 
than  the  red  rays  or  shorter  than  the  violet  rays,  are  called  the 
visible  rays.  It  can  be  easily  proved  that  white  light  contains 
other  rays,  some  of  which  extend  beyond  the  red  and  others 
beyond  the  violet;  in  other  words,  some  of  these  rays  are  less 
refrangible  than  the  red  rays  and  others  are  more  refrangible 
than  the  violet  rays.  These  rays  are  called  the  invisible  rays. 
The  invisible  rays  beyond  the  red  are  frequently  called  the 
ultra-red  rays,  those  beyond  the  violet  the  ultra-violet  rays, 
ultra  meaning  beyond. 

The  ultra-red  rays  are  heat  rays,  and  their  presence  may 
be  proved  by  placing  a  thermometer  in  that  part  of  the  spectrum 
beyond  the  red,  when  a  rise  in  temperature  will  be  observed. 
The  ultra-violet  rays  are  actinic  or  chemical  rays,  and  their 
presence  is  revealed  by  their  action  on  the  photographic  plate. 

158.  Primary  Colors. — Of  the  seven  principal  colors  of  the 
spectrum,  three — red,  yellow,  and  blue — are  called  the  primary 
colors,  because  by  properly  mixing  these  colors,  any  other  color, 
hue,  shade,  or  tint  may  be  obtained,  including  white. 

159.  Secondary  and  Tertiary  Colors. — When  two  primary 
colors  are  mixed,  the  result  is  a  secondary  color;  thus,  red  and 
yellow  produce  orange,  yellow  and  blue  produce  green,  and  red 
and  blue  produce  violet.  The  secondaries  vary  in  shade  and 
tint  in  accordance  with  the  proportions  of  the  primaries  used. 

When  two  secondary  colors  are  mixed,  the  result  is  called  a 
tertiary  color.  A  tertiary  color  is  said  to  be  a  combination  of 
four  colors,  because  the  two  secondaries  must  have  one  color  in 
common,  which  is  counted  as  two  colors,  and  these  added  to  the 
two  primaries  make  four  colors;  thus,  green  and  orange  produce 
brown  =  yellow  +  blue  +  yellow  +  red.  Hence,  brown  may 
have  about  twice  as  much  yellow  as  it  has  of  red  or  blue,  according 
to  the  proportions  used. 

160.  Natural  Colors. — The  color  of  an  object  as  seen  in  a  clear 
white  light  is  called  its  natural  color.  The  reason  that  an  object 
has  color  is  because  it  absorbs  all  the  visible  light  rays  except 
those  that  are  necessary  to  produce  the  natural  color,  the  latter 
being  reflected  to  the  eye.  This  fact  is  easily  proved:  Place  a 
red  rose  in  the  path  of  the  beam  of  light  between  the  slit  ab  and 
the  prism  P  in  Fig.  38;  the  rose  will  be  red.     If  held  against  the 


112  ELEMENTS  OF  PHYSICS  §4 

screen  in  the  red  part  of  the  spectrum,  it  will  still  be  red;  but  if 
placed  in  the  yellow  part  or  any  part  above  the  yellow,  it  will 
appear  black,  showing  that  it  has  absorbed  the  light  rays,  none  of 
them  being  reflected.  In  other  words,  a  red  object  reflects  only 
red  rays,  a  yellow  object  only  yellow  rays,  and  a  blue  object  only 
blue  rays.  A  green  object  reflects  both  yellow  and  blue  rays, 
and  these  combine  to  produce  green,  etc. 

An  object  never  appears  in  its  natural  color  or  colors  unless 
viewed  in  a  clear  white  light  or  in  a  light  having  the  exact  color 
of  the  object.  It  is  for  this  reason,  that  it  is  practically  impossible 
to  match  colors,  shades,  etc.  under  artificial  light,  because  no 
artificial  light  has  yet  been  produced  that  is  a  pure  white.  A 
white  lily  appears  white  in  a  white  light,  because  it  reflects  all  the 
component  colors  of  white  light;  but  if  it  be  placed  in  a  red 
light,  it  will  appear  to  be  red,  and  if  placed  in  a  blue  light,  it 
will  appear  to  be  blue,  etc.,  reflecting  in  each  case  the  rays  that 
make  up  the  light. 

The  color  of  a  substance,  as  a  sample  of  paper,  can  be  re- 
presented numerically  by  means  of  the  tint  photometer,  which 
measures  the  percentage  of  red,  yellow  and  blue  in  light  re- 
flected from  the  surface.  This  instrument  is  described  in  the 
Section  on  "paper  testing". 

161.  The  Spectroscope;  Spectrum  Analysis. — Any  substance 
— solid,  liquid,  or gaseous — that  can  be  burned,  giving  off  a  gaseous 
flame,  has  a  spectrum  peculiar  to  the  substance  and  characteris- 
tic of  it.  These  spectrums  have  bright  lines,  the  number  of 
which  and  their  color  and  position  identify  the  substances. 
The  spectrum  of  burning  sodium  shows,  in  addition  to  other  lines, 
two  conspicuous,  bright  yellow  lines  that  are  so  close  together  as 
to  appear  almost  as  one  line;  consequently,  whenever  these  two 
lines  appear  in  a  spectrum  of  a  flaming  substance,  and  in  their 
proper  position,  it  is  certain  that  sodium  is  present  in  the  substance. 
The  flame  of  carbon  shows  two  distinct  lines,  one  green  and  the 
other  indigo.  Other  substances  have  their  own  special  character- 
istics, which  have  been  examined  and  tabulated,  and  when  burned 
may  be  identified  by  their  spectrums.  This  method  of  identify- 
ing substance  is  called  spectrum  analysis;  by  means  of  it,  the 
composition  of  the  sun  and  the  stars  has  been  ascertained. 

162.  The  spectroscope  is  an  instrument  used  in  determining 
the  color  and  position  of  the  bright  lines  that  are  peculiar  to  the 


§4 


LIGHT 


113 


substance  being  analyzed.  It  consists  of,  see  (a),  Fig.  39,  a 
prism  P  enclosed  in  a  case  from  which  light  is  excluded  and  to 
which  are  attached  three  tubes  M,  N,  and  T.     Light  from  the 


burning  substance  enters  tube  M  through  a  very  narrow  slit  s, 
passes  through  the  lens  V,  which  makes  the  rays  parallel,  then 
passes  through  the  prism  P,  which  disperses  the  rays,  resolves 
them  into  their  spectral  colors,  and  refracts  them  through  the 


114  ELEMENTS  OF  PHYSICS  §4 

tube  T  to  the  eye.  Before  reaching  the  eye,  they  pass  through  the 
lens  I",  which  converges  them  to  a  focus,  beyond  which  they  are 
magnified  by  the  eyepiece,  the  tube  T  being  a  telescope.  The 
tube  N  contains  a  scale  at  r,  the  light  rays  from  which  pass 
through  the  lens  V",  which  makes  the  rays  parallel;  they  strike 
the  prism  P,  and  are  reflected  through  the  telescope  tube  T  to  the 
same  relative  position  as  the  magnified  image  of  the  spectrum 
received  from  the  rays  passing  through  the  slit  s,  thus  making 
it  possible  to  measure  with  a  high  degree  of  precision  the  position 
of  any  bright  lines  that  may  occur  in  the  spectrum  of  the  sub- 
stance being  analyzed.  A  perspective  view  is  shown  at  (6), 
Fig.  39.  As  may  be  supposed,  there  are  many  makes  of  spectro- 
scopes, the  simpler  ones  having  no  scale  for  measuring  the  posi- 
tions of  the  bright  lines. 

By  means  of  spectrum  analysis,  several  of  the  chemical  ele- 
ments have  been  discovered,  notably  helium,  which  was  found  in 
the  sun  and  named  before  it  was  discovered  as  one  of  the  con- 
stituents of  the  earth.  In  some  cases,  spectrum  analysis  can  be 
relied  on  when  chemical  analysis  fails;  thus,  human  blood  and  the 
blood  of  a  pig  cannot  be  distinguished  chemically,  but  they  can  be 
identified    separately    by    the    spectroscope. 

163.  Complementary  Colors. — If  two  colors  when  mixed  pro- 
duce white  they  are  said  to  be  complementary.  For  instance, 
if  a  line  be  drawn  across  the  spectrum  of  white  light,  and  one  of 
the  two  colors  contains  all  the  colors  on  one  side  of  the  line  and  the 
other  color  contains  all  the  colors  on  the  other  side  of  the  line, 
the  two  colors  when  mixed  will  produce  white  and  are  called 
complementary. 

The  following  interesting  experiment  will  show  the  comple- 
mentary colors:  Place  on  a  black  surface  a  small  square  or  disk 
of  some  bright  color,  as  red,  yellow,  or  blue,  and  gaze  at  it  steadily 
for  about  a  minute;  then  look  at  a  white  wall  or  white  surface  of 
any  kind,  as  a  sheet  of  paper,  and  the  outline  of  the  surface 
looked  at  will  immediately  appear,  but  its  color  will  be  the  com- 
plement of  that  of  the  object.  If  the  object  is  red,  the  comple- 
ment on  the  white  surface  will  be  bluish-green,  if  yellow,  it  will  be 
blue,  and  if  green,  it  will  be  crimson,  etc.  The  following  are 
some    of    the    colors    and    their    complements: 

Red  Orange       Yellow  Violet  Green 

Bluish-green     Greenish-blue     Blue     Greenish-yellow     Crimson 


§4  LIGHT  115 

The  explanation  of  this  phenomenon  is  that  the  nerves  of  the 
eye  which  respond  to  the  color  of  the  object  gazed  at  become 
fatigued  and  do  not  respond  to  the  white,  which  includes  all  colors, 
until  a  short  time  after  the  gaze  has  been  removed  from  the  object; 
the  remaining  nerves,  which  respond  instantly  to  the  other  colors, 
blend  these  colors  and  produce  the  complementary  color.  As 
will  be  naturally  expected,  the  experiment  will  be  more  successful 
if  the  object  is  placed  on  a  black  surface,  since  the  only  rays  re- 
flected to  the  eye  will  then  be  those  from  the  object. 

164.  Lengths  and  Vibrations  of  Light  Waves. — As  was  previ- 
ously mentioned,  the  wave  length  and,  consequently,  the  number 
of  vibrations  per  second  varies  in  accordance  with  the  color.  It  is 
the  rate  of  vibration  that  determines  any  particular  color.  The 
wave  lengths  have  been  measured  for  different  colors,  and  know- 
ing the  wave  length  for  any  color,  the  rate  of  vibration  for  that 
color  may  be  found  by  dividing  the  velocity  of  light  by  the  length 
of  the  wave.     Thus,  the  wave  length  of  a  red  ray 

Wave  length  in                      Vibration  per 

inches  seconds 

Red 0000268  441,000,000,000,000  =  441  X  1012 

Orange 0000248  476,000.000,000,000  =  476  X  1012 

Yellow 0000228  518,000,000,000,000  =  518  X  10'2 

Green 0000204  579,000,000,000,000  =  579  X  1012 

Blue 0000182  649,000,000,000,000  =  649  X  1012 

Indigo 0000175  675,000,000,000.000  =  675  X  1012 

Violet 0000166  712,000,000.000,000  =  712  X  1012 

is  .0000268  inch;  the  velocity  of  light  is  186,400  miles  =  186400 
X 63360  inches  per  second;  hence,  the  number  of  vibrations  per 
second  required  to  make  light  visible  as  a  red  ray  is 

^Sf^= ^.000,000,000 

which  may  be  written  441  X  1012,  a  number  almost  inconceiv- 
ably large.  The  wave  lengths  and  the  vibrations  per  second  for 
the  seven  principal  colors  are  given  in  the  foregoing  table. 

165.  Mixing  Pigments. — Any  coloring  material  used  for  paint- 
ing or  printing  is  called  a  pigment.  Coloring  materials  used  in 
the  paper  industry  arc  usually  dye-stuffs,  which,  when  fixed  on  the 
fiber  have  the  properties  of  pigments.  Mixing  pigments  to  get  a 
certain  desired  color  is  quite  a  different  problem  from  that  of 
mixing  or  blending  colored  lights  to  get  the  same  color.  For 
instance,  when  yellow  light  is  added  to  a  particular  shade  of  blue, 


116  ELEMENTS  OF  PHYSICS  §4 

the  result  is  white  light,  because,  according  to  Art.  163,  the  two 
colors  are  complementary.  If  a  yellow  pigment  be  added  to  a 
blue  one,  the  resulting  color  will  be  green;  the  reason  for  this  is 
that  the  yellow  pigment  absorbs  the  blue  and  violet,  the  blue 
pigment  then  absorbs  the  red  and  yellow,  with  the  result  that  only 
the  green  is  left  to  be  reflected.  The  final  result  will  be  obtained 
regardless  of  which  color  is  applied  first;  simply  allow  it  to  dry 
and  then  apply  the  other  color  on  top  of  the  first  Exact  re- 
productions of  the  spectral  colors  cannot  be  obtained  with  pig- 
ments, because  the  pigments  themselves  do  not  have  exactly  the 
same  shades  as  the  colors  of  the  spectrum,  called  the  spectral 
colors. 

166.  Three-color  Process. — What  is  called  the  three-color 
process  in  printing  is  a  more  or  less  successful  attempt  to  reproduce 
objects  in  their  natural  colors  when  using  only  three  colors  of  ink. 
When  the  area  of  a  disk  is  divided  into  three  sectors,  and  one 
sector  is  colored  red,  the  second  green,  and  the  third  blue-violet, 
and  the  disk  is  caused  to  revolve  rapidly,  the  three  colors  will 
blend  into  one  single  color.  The  nature  of  the  color  obtained  will 
depend  upon  the  relative  areas  of  the  sectors,  and  by  varying 
these,  any  desired  color  may  be  obtained.  Calling  the  three 
colors  just  mentioned  the  primary  colors,  the  primary  pigments 
are  the  complements  of  these,  and  are  (in  order)  peacock  blue, 
crimson,  and  light  yellow.  It  is  to  be  observed  that  when  the 
three  primary  colors  mentioned  above  are  mixed,  the  result  will 
be  white;  but  when  the  three  primary  pigments  are  mixed,  the 
the  result  is  black,  because  they  absorb  all  the  colors  of  white 
light.     If    black    is    mixed    with    white,    the    result    is    gray. 

In  printing,  the  three  primary  pigments  are  applied  to  white 
paper  in  the  following  manner:  Three  different  photographs  of 
the  object,  which  may  be  a  natural  object  or  a  painting,  are  made, 
a  gelatine  screen,  transparent,  of  the  same  color  as  one  of  the 
primary  colors  being  placed  in  front  of  the  camera  lens,  a 
different  color  for  each  photograph,  and  care  is  taken  to  have  each 
photograph  exactly  the  same  in  size.  Then  halftone  blocks  are 
made  in  the  usual  way  from  these  photographs.  The  colored 
print  is  then  made  by  printing  on  white  paper  from  one  of 
these  halftone  blocks,  using  an  ink  that  is  of  the  same  color  as  the 
complement  of  the  color  of  the  screen  that  was  used  when  the 
photograph  was  taken.  After  the  ink  has  dried,  the  sheet  is  run 
through  the  press  again,  one  of  the  other  halftone  blocks  being 


H&U 

_» <L 

(E?33SfeSr«SS 

K"vH 

rm 

^BPg 

■4 

2  1  J1  -^  C  J^-sT7 

Fig.  40 


ELEMENTS  OF  PHYSICS 

(PART  2) 


EXAMINATION  QUESTIONS 

(1)  Steam  is  cut  off  in  an  engine  cylinder  at  fths  stroke  and 
expands  to  the  end  of  the  stroke.  Assuming  that  the  fall  in 
pressure  follows  Boyle's  law  for  increase  in  volume  (which  is 
approximately  true),  what  will  be  the  pressure  at  the  end  of 
the  stroke,  if  the  pressure  at  cut  off  is  126  lb.  per  sq.  in.  gauge? 

Ans.  38+  lb.  per  sq.  in.  gauge. 

(2)  If  4.6  cu.  ft.  of  air  at  96°F.  expand  at  constant  pressure 
until  the  temperature  becomes  53°F.,  what  will  be  the  volume? 

Ans.  4.244 -cu.  ft. 

(3)  A  vessel  holding  1.58  cu.  ft.  is  filled  with  air  at  a  pressure 
of  14.65  lb.  per  sq.  in.  abs.;  if  the  temperature  of  the  air  is 
70°F.  and  it  is  heated  to  700°F.,  what  will  be  its  pressure? 

Ans.  32.08  lb.  per  sq.  in.  abs. 

(4)  If  2.66  cu.  ft.  of  air  at  62°F.  and  a  pressure  of  1  atmosphere, 
is  compressed  to  .52  cu.  ft.,  what  will  be  the  tension  when  the 
air  has  a  temperature  of  112°F.? 

Ans.  67.71  lb.  per  sq.  in.  gauge. 

(5)  Taking  the  specific  gravity  of  nitrogen  as  .971,  what  is 
the  weight  of  875  cu.  ft.  when  the  tension  is  20  lb.  per  sq.  in. 
abs.  and  the  temperature  is  80CF.?  Ans.  85.186  lb. 

(6)  Referring  to  the  last  example,  what  is  the  weight  of  1  cu. 
ft.  of  nitrogen  at  32°F.  and  a  tension  of  1  atmosphere? 

Ans.  .078517  lb. 

(7)  Referring  to  Questions  5  and  6,  if  3.8  cu.  ft.  of  nitrogen  at 
60°F.  and  a  tension  of  1  atmosphere  is  heated  at  constant  volume 
to  1000°F.,  (a)  how  many  B.t.u.  must  be  expended?  (b)  how  many 
foot-pounds  of  work  are  equivalent  to  this?  (c)  what  will  be  the 
tension?  r  (a)  46>7  B  t  u 

Ans.       (6)  36,333  ft.-lb. 

1  (c)   41. 3 -lb.  per  sq.  in.  abs. 
§4  119 


120  ELEMENTS  OF  PHYSICS  §4 

(8)  If  23  pounds  of  water  at  45°  are  mixed  with  18  pounds  at 
190°  and  a  piece  of  ice  weighing  3  pounds  is  placed  in  the  mix- 
ture, what  will  be  the  temperature  after  the  ice  has  melted  and 
the  entire  mixture  has  the  same  temperature,  not  considering 
the  vessel  holding  it?  Ans.  93.61°. 

(9)  A  platinum  ball  weighing  15.708  oz.  is  heated  to  a  tem- 
perature of  2700°F.;  it  is  then  placed  in  a  wrought-iron  vessel 
containing  2  lb.  5  oz.  of  water.  If  the  temperature  of  the 
vessel  and  the  water  is  70°F.,  what  will  be  the  temperature  of 
the  mixture?  Ans.  104.3°F. 

(10)  Referring  to  the  last  question,  suppose  the  temperature 
of  the  ball  had  not  been  known,  but  the  temperature  of  the  mix- 
ture had  been  found  to  be  104°F.,  what  would  be  the  temperature 
of  the  ball?  Ans.  2677°F. 

(11)  How  many  gram  calories  are  equivalent  to  (a)  2.571 
B.t.u.?  (6)  to  1  B.t.u.?  Ans.    f  (a)  647.9  gram  cal. 

I  (6)  252  gram  cal. 

(12)  What  is  the  temperature  of  saturated  steam  in  a  soda 
pulp  digester  when  the  pressure  is  110  lb.  per  sq.  in.  gauge? 

Ans.  344.1°F. 

(13)  What  is  the  total  heat  of  1  pound  of  steam  in  a  condenser 
when  the  vacuum  gauge  reads  11.5  in.?  Ans.  1141.2  B.t.u. 

(14)  What  is  the  total  heat  of  7  lb.  of  steam  when  the  pressure 
is  60  lb.  per  sq.  in.  gauge?  Ans.  8270  B.t.u. 

(15)  What  is  (a)  a  standard  candle?  (b)  what  is  the  candle- 
power  of  a  lamp  that,  when  placed  112  in.  from  a  screen,  illu- 
minates the  screen  with  the  same  intensity  as  a  standard  candle 
at  a  distance  of  24  in.?  Ans.  27|  c.p. 

(16)  What  is  meant  (a)  by  diffusion  of  light?  (b)  by  refraction 
of  light?  (c)  by  reflection  of  light?  (d)  what  is  the  relation  be- 
tween the  angle  of  incidence  and  the  angle  of  reflection? 

(17)  What  are  (a)  the  primary  colors?  (6)  what  is  meant  by 
secondary  and  tertiary  colors?  (c)  what  causes  color? 

(18)  What  (a)  is  a  pigment?  (6)  If  three  pigments  having 
the  colors  of  the  three  primary  colors  are  mixed,  what  is  the 
result?  (c)  three  beams  of  light  having  the  colors  of  the  three 
primary  colors  are  mixed,  what  is  the  result?  (d)  What  is  the 
cause  of  the  difference  in  the  two  results? 


INDEX 


Note — The  paging  begins  with  1  in  each  section,  and  each  section  has  its  number  printed 
mi  the  inside  edge  of  the  headline  of  each  page.  To  find  a  reference,  as  "Abbreviations  on 
drawings,  §3,  pl6,"  glance  through  the  volume  until  §3  is  found  and  then  find  page  16. 


Abbreviation  in  division,  52,  p31 

in  multiplication,  §2,  p30 
Abbreviations  on  drawings,  §3,  pl6 
Absolute  index  (of  refraction),  §4,  pl03 

pressure,  definition  of,  §4,  p53 

temperature,  §4,  p61 

zero  of  pressure,  §4,  p54 

zero  of  temperature,  §4,  p61 
Abstract  number,  §1,  p2 
Acceleration,  definition  of,  §4,  p8 

positive  and  negative,  §4,  p9 

unit  of,  §4,  p9 
Accuracy  in  calculation,  §2,  pp27-33 

in  measurements,  §2,  p28 

in  numerical  operations,  §2,  p29 
Actinic  rays,   §4,  p91 
Action,  line  of,  §4,  pl9 
Acute  angle,  {2,  p46 
Addition,  definition  of,  {1,  pll 

of  compound  numbers,  §1,  pl20 

of  decimals,  §1,  p68 

of  fractions,  rule  for,  §1,  p55 

of  mixed  numbers,  §1,  p56 

of  monomials,  §2,  p4 

of  numbers,  $1,  ppl3-18 

of  polynomials,  §2,  plO 

rule  for,   §1,  pi  5 

sign  of,  §1,  pll 

table  of,  §1,  pi  2 
Adhesion,  definition  of,  54,  plO 
Adjacent  angles,  52,  p45 
Aggregation,  signs  of,  5L  p76;  52,  pl5 
Air,  definition  of,  54,  p45 
Altitude  of  cone,  52,  pl21 

of  cylinder,  52,  pill 

of  parallelogram,  52,  p61 

of  prism,  52,  pl02 

of  pyramid,  52,  pi 04 

of  triangle,  52,  p50 
Amount,  definition  of,  51.  plOl 

rule  to  find.  51,  plOl 
Amplitude,  definition  of,  54.  p91 
Analysis,  spectrum,  54,  pi  12 
Aneroid  barometer,  description  of,  54,  p*>2 
Angle,  acute  and  obtuse,  52,  p46 

critical,  for  emergent  ray,  §4,  pl03 

definition  of,   §2,  p45 

inscribed,  §2,  p71 

of  deviation,  §4,  plOl 


Angle  of  incidence,  54,  pp97,  101 

of  reflection,  54,  p97 

of  refraction,  54,  plOl 

re-entrant,  52,  p66 

right,    definition    of,    51,    ppllO,    112; 
52,  p46 

sides  of,  52,  p45 

vertex  of,  52,  p45 
Angles,  adjacent,  52,  p45 

circular  measure  of,  52,  p83 

equality  of,  52,  p54 

measurement  of,  52,  p46 
Angular  measure,  table  of,  §1,  pi  10 
Apothem,  definition  of,  52,  p67 
Arabic  notation,  §1,  pp3,  5 
Arc,  length  of  ciicular,  52,  p85 

(of  circle),  definition  of,  52,  p70 

of  curve,  52,  p44 
Arc,  definition  of,  §1,  pi  14 
Area,  entire,  definition  of,  52,  pl03 

of  any  plane  figure,  52,  p93 

of  circle,  to  find,  52,  p77 

of  cylinder,  52,  pplll,  112 

of  ellipse,  52,  p92 

of  fillet,  52,  p88 

of  frustum  of  cone,  52,  pi 23 

of  parallelogram,  §2,  p62 

of  prism,  52,  pl03 

of  pyramid,  52,  pl05 

of  regular  polygons,  52,  p68 

of  ring,  52,  p82 

of  sector  and  segment,  52,  p87 

of  sphere,  52,  pl25 

of  zone,  52,  pl26 
Arithmetic,  definition  of,  51.  p2 
Arithmetical  mean,  51,  pl25 
Application,  point  of,  54,  pl4 
Atmosphere,  definition  of,  54,  p45 

pressure  of,  54,  p46 

value  of,  54,  p46 
Atom,  definition  of,  54,  p2 
Attraction,  capillary,  54,  p39 
Average  of  numbers,  Jl,  pl25 
Avoirdupois  weight,  table  of,  51.  pl08 
Axes,  major  and  minor,  of  ellipse,  52,  p91 
Axis  of  cone,  52,  pl20 

of  pyramid,  52,  pl05 

of  sphere,  52,  pi  24 

of  symmetry,  52,  pl37 

principal,  of  lens,  §4,  pl06 


121 


122 


INDEX 


B 

Back  view,  §3.  p7 
Barometer,  definition  of,  §4,  p51 
Barometers,  mercurial  and  aneroid,  54,  p52 
Base,  definition  of,  §1,  pp98,  100 

of  parallelogram,  j2,  p62 

of  spherical  sector,  §2,  pl27 

of  triangle,  §2,  pol 

rule  to  find,  |1,  pplOO,  101,  102 
Bases  of  a  solid,  §2,  plOl 

of  cylinder,  §2,  pi  10 

of  spherical  segment,  §2,  pl26 
Beam  of  light,  rays,  §4,  p92 
Beaum6  scale  for  hydrometer,  §4,  p37 
Belt,  to  find  length  of,  ppSO,  81 
Binomial,  definition  of,  §2,  p9 
Blow,  definition  of,  §4,  pl9 
Body,  definition  of,  §4,  pi 
Boiling  point,  table  of,  54,  p86 
Bore  or  bored,  definition  of,  §3,  pl6 
Bottom  view,  §3,  p" 
Boyle's  law,  §4,  p57 
Brightness  of  light,  §4,  p95 
British  imperial  gallon,  §1,  pl09 

thermal  unit,  definition  of,  §4,  p67 
Brittleness,  definition  of,  §4.  pG 
Broken  and  dotted  lines,  use  of,  §3,  pl2 

line,  §2,  p44 

lines,  use  of,  §3,  pl2 
Bunsen's  method  of  measuring  intensity  of 

light,  §4,  p96 
Buoyancy,  definition  of,  §4,  p.12 
Bushel,  British,  §1,  pi  10 

Winchester,  §1,  pl09 


Calculation,  accuracy  in,  §2,  pp27-33 

definition  of,  §1.  p2 
Calorie,  definition  of,  §4,  p67 

gram  or  small,  §4,  p68 

kilogram  or  large,  §4,  p68 
Cancelation,  §1,  p37 
Candlepower,  definition  of,  §4,  p95 
Candle,  standard,  §4,  p95 
Capacity,  heat,  §4,  p68 
Capillarity,  definition  of,  §4.  p39 
Capillary  attraction,  §4,  p39 

attraction,  examples  of,  §4,  p41 
Center  of  gravity,  §2,  pl40 

of  gravity,  to  find  experimentally,   $2, 
pl41 

lines,  $3,  pplO,  13 

of  ellipse,  §2,  p91 

of  polygon,  geometrical,  §2,  pGG 

of  sphere,  §2,  pi 24 

of  symmetry,  §2,  ppl38,  140 

optic,  §4,  plOO 
Centigrade  degrees  to  Fahrenheit,  §4,  p44 

scale,  §4,  p43 
Central  angle,  {2,  p71 


C.  G.  S.  system,  §4,  pl2 
Chain  discount,  $1,  pl03 
Charles'  law,  §4,  p60 
Check  for  division,  §1,  p32 

for  multiplication,  §1,  p26 
Chord,  definition  of,  §2,  p70 

of  arc,  to  find,  §2,  p75 

of  half  the  arc,  to  find,  $2,  p76 
Cipher,  naught,  or  zero,  §1,  po 
Circle,  definition  of,  §2,  p68 

to  describe  a,  §2,  p69 

to  find  area  and  circumference,  {2,  p77 

to  find  diameter  or  radius,  §2,  p77 
Circles,  concentric  and  eccentric,  §2,  p81 

great  and  small,  52,  pl25 

properties  of,  §2,  pp71-74 
Circular  arc,  definition  of,  §2,  p70 

arc,  length  of,  §2,  p85 

measure  of  angles,  §2,  p83 
Circumference,  definition  of,  §2,  p70 

of  circle,  to  find,  §2,  p77 

of  ellipse,  §2,  p92 
Circumscribed  polygon,  §2,  pS9 
Clearing  equation  of  fractions,  §2,  p20 
Coefficient,  definition  of,  52,  p3 

of  conductivity,  definition  of,  §4,  p76 

of  conductivity,  table  of,  §4,  p70 

of  expansion,  §4,  p72 
Coefficients  of  expansion,  table  of,  §4,  p73 
Cohesion,  definition  of,  §4,  p9 
Collecting  terms,  definition  of,  §2,  p!9 
Coloring  paper,  reason  for,  §4,  pi  18 
Color,  natural,  §4,  pill 
Colors,  complementary,  §4,  pi  14 

primary,  §4,  pill 

prismatic,  §4,  p]09 

secondary  and  tertiary,  §4,  pill 
Combined  pressures  on  submerged  surface, 

§4,  p29 
Common  denominator,  §1,  po2 

divisor,  greatest,  $1,  p43 

multiple,  least,   §1,  p45 
Compasses,  definition  of,  §2,  p69 
Complementary  colors,  §4,  pi  14 
Complete  divisor.  §1,  p93 
Complex  fraction,  §1,  p62 
Composite  number,  definition  of,  §1,  p35 
Compound  fraction,  §1,  p64 

proportion,  §1,  p83 

microscope,  §4,  pl08 

number,  definition  of,  §1,  pl06 

numbers,  addition  of,  §1,  pl20 

numbers,  division  of,  §1,  pl23 

numbers,  multiplication  of,  §1,  pl22 

numbers,  reduction  of,  §1,  ppl  16-120 

numbers,  subtraction  of,  §1 ,  pi 21 
Compressibility,  definition  of,  §4,  p4 
Computation,  definition  of,  §1,  p2 
Concave  lens,  §4,  pl06 

polygon,  J2,  p66 
Concavo-convex  lens,  54,  pl06 


INDEX 


123 


Concentric  circles,  §2,  p81 

Concrete  number,  §1,  p2 

Conditions,  standard,  definition  of,  J41  p36 

Conduction  of  heat,  §4,  p75 

Conductivity,    coefficient  of,    table   of,    §4, 

p76 
Conductors  of  heat,  good  and  poor,  §4,  p75 
Cone,  area  and  volume  of,  §2,  pi 21 

definition  of,  §2,  pl20 

frustum   of,    area   and    volume    of,    §2, 
pl23 

of  light  rays,  §4,  p92 

of  revolution,  §2,  pl20 

right,  §2,  pl20 

slant  height  and  altitude  of,  §2,  pl21 

vertex  and  axis  of,  §2,  pi 20 
Conical  surface,  §2,  pl20 
Constants,  definition  of,  §2,  p32 
Constant  velocity,  §4,  p8 
Convection,  definition  of,  §4,  p77 
Conventional  sections,  §3,  p22 
Convex  area  of  cylinder,  §2,  pill 

lens,  §4,  pl06 

polygon,  §2,  p66 
Cord,  definition  of,  §1,  pl08 
Cored,  definition  of,  §3,  pi 6 
Critical  angle  for  emergent  ray,  §4,  pl03 
Cross  hatching,  §3,  pl7 

section,  §2,  pl02;  §3,  23 
Crossed  belt,  to  find  length  of,  §2,  p81 
Crown  of  pulley,  §3,  p23 
Cylinder,  altitude  of,  §2,  pill 

bases  of,  §2,  pi  10 

convex  area  of,  52,  pill 

definition  of,  §2,  ppllO,  117 

development  of,  §2,  pill 

diameter  of,  §2,  pill 

of  revolution,  §2,  pi  10 

projection  of,  §3,  p25 

right,  §2,  pi  10 

surface  of,  §2,  pi  10 

volume  of,   §2,  pll4 
Cylindrical  ring,  volume  of,  §2,  pl42 
Cube  root  of  numbers,  §2,  p33-39 
Cubes  and  fifth  powers,  table  of,  §2,  p35 
Cubical  contents,  definition  of,  §2,  pl03 
Cubic  centimeter,  §1,  pi  15 

foot,  definition  of,  §1,  pill 

measure,  table  of,  §1,  pi 08 
Curve  or  curved  line,  §2,  p44 

D 

d.,  dia.,  or  diam.,  definition  of,  §3,  plG 
Decimal  fractions,  §1,  p67 

point,  definition  of,  §1,  p8 
Decimals,  addition  and  subtraction  of,    §1, 
p68 

definition  of,  §1,  ppS,  07 

division  of,  rule  for,   §1,  p73 

multiplication  of,  §1,  pC9 


Decimals,  reading,  §1,  p9 

reduction  of,  to  common  fractions,   {1, 
pp67,  68 

to  reduce  to  fractions  having  a  given 
denominator,  §1,  p75 
Decomposition  of  light,  §4,  pi  10 
Degree  of  equation,  §2,  p23 
Degrees  (of  arc),  definition  of,  §1,  pllO 

on  hydrometer  scale,  §4,  p37 

on  thermometer  scale,  §4,  p43 
Denominate  number,  definition  of,  §1,  plOC 
Denominator,  common  and  least  common, 
§1,  p52 

of  a  fraction,  §1,  p48 

reduction    of   fraction    to   a   given,    §1, 
p51 
Density    and    specific    gravity,     difference 
between,  §4,  p35 

definition  of,  §4,  pl4 
Development  of  cylinder,  §2,  pill 
Deviation,  angle  of,  §4,  plOl 
Dew  point,  §4,  p87 
Diagonal  of  parallelogram,  §2,  pOl 

of  square,  length  of,  §2,  p63 
Diameter  of  circle,  definition  of,  §2,  p70       # 

of  circle,  to  find,  §2,  p77 

of  cylinder,  §2,  pill 

of  sphere,  §2,  pi 24 
Diameters  of  ellipse,  §2,  p92 
Difference,  definition  of,  §1,  pl8 

(in  percentage)  definition  of,  $1,  plOl 

rule  to  find,  §1,  pl02 
Diffusion  of  light,  §4,  plOO 
Digits,  §1,  po 
Dimension  lines,  §3,  pl3 
Dimensions,  over-all,  §3,  pl4 
Direct  ratio,  §1,  p78 
Direction  of  motion,  §4,  p7 
Directrix,  definition  of,  §2,  pll7 
Discount,  chain,  §1,  pl03 

definition  of,  §1,  pl02 

rate,  rule  for  equivalent,  §1,  pl03 
Dispersion  of  light,  §4,  pi  10 
Dissolve,  definition  of,  §4,  p2 
Dividend,  definition  of,  §1,  p28 

trial,  §1,  p92 
Divisibility,  definition  of,  §4,  p2 

of  numbers,  §1,  p36 
Divisible,  definition  of,  §1,  p35 
Division,  abbreviation  in,  §2,  p31 

by  a  power  of  10,  §1,  p07 

check  for,  §1,  p32 

definition  of,  §1,  p27 

long,  §1,  p31 

of  compound  numbers,  §1,  pl23 

of  decimals,  rule  for,  §1,  p73 

of  fractions,  rule  for,  §1,  pG2 

of  monomials,  §2,  p8 

of  polynomials,  §2,  ppl3-15 

rule  for,  §1,  p33 

short,  $1,  p29 


124 


INDEX 


Division,  signs  of,  §1,  p28 
Divisor,  complete,  §1,  p93 

definition  of,  $1,  p28 

greatest  common,  §1,  p43 

trial,  U,  p92 
Dollar,  definition  of,  §1,  pi  12 
Dotted  lines,  use  of,  §3,  pl2 
Double-concave  lens.  §4.  pl06 
Double  sign,  meaning  of,  52,  p24 
Dozen,  definition  of,  §1,  pi  13 
Drawing,  definition  of,  §3,  p2 

perspective,  §3,  p2 

projection,   §3,  p2 

scale  of,   §3,  pl5 

scaling  a,   §3,  pi 6 

to  scale,  §3,  pl4 

working,   §3,  p8 
Drawings,  abbreviations  on,   §3,  pl6 

notes  on,  §3,  pl6 

reading,  directions  for,   §3,  pp25-2S 
Dry  measure,  table  of,  §1,  pl09 

vapor,  J4,  p83 
Ductility,  definition  of,  §4,  p6 


E 


Eccentric  circles,  §2,  pSl 
Edges  of  a  solid,  §2,  plOl 
Elasticity,  definition  of,  §4,  p4 
Element,  definition  of,  §2,  ppllO,  120 
Elevation,  front,   §3,  pC 

rear,  53,  p7 

side,  §3,  p6 
Ellipse,  definition  of,  §2,  p91 

diameters  of,  §2,  p92 

foci  of,  §2,  p91 

periphery  and  area  of,  £2,  p92 

vertexes  of  and  center  of,  §2,  p91 
Energy,  definition  of,  §4,  pl6 

potential  and  kinetic,  §4,  plC 

total,  54,  P17 
Entire  area,  definition  of,  §2,  pl03 
Equality  sign,  definition  of,  §1,  pll 
Equation,  clearing,  of  fractions,  §2,  p20 

definition  of,  $2,  plS 

degree  of,  §2,  p23 

members  of,  $2,  plS 

quadratic,  formula  for  roots  of,  §2,  p25 

roots  of,  §2,  p25 

transposing  terms  of,   §2,  pl9 
Equations,   identical  and   independent,    52. 
pl8 

linear,  §2,  p24 

of  first  degree,  §2,  p24 

of  second  degree,  §2,  p24 

quadratic,  §2,  p23 

quadratic,  solution  of,  §2,  p25 

transformation  of,  §2,  pl9 
Equilibrium,  definition  of,  §4,  pl8 
Equivalent  discount  rate,  rule  for,  [1,  pl03 
Ether,  definition  of,  §4,  p78 


Even  number,  definition  of,  51,  p35 
Expansibility,  definition  of,  $4,  p4 
Expansion,  coefficient  of,  {4,  p72 

linear,  surface  and  cubic,  54,  p72 
of  bodies  by  heat,  54,  p72 
Exponent,  definition  of,  §1,  p65 
Exponents,  literal  and  numerical,  52,  p4 
Extension,  definition  of,  54,  pi 


f.  or  fin.,  definition  of,  53,  plG 
Faced,  definition  of,  53,  pi  7 
Faces  of  lens,  §4,  plOG 

of  solid,  52,  plOl 
Factors,  definition  of,  §1,  p22 

prime,  |1,  p36 
Fahrenheit  degrees,  to  centigrade,  54,  p44 

scale,  54.  p43 
Fifth  powers  and  cubes,  table  of,  52,  p35 

root  of  numbers,  52,  p33-39 
Figure,  order  of,  $1,  p6 
Figures,  definition  of,  §1,  p5 

significant,  52,  p27 

significant,  number  required,  52,  p32 

similar,  52,  pl31 

symmetrical,  52,  pi 37 
Fillet,  area  of,  52,  p88 
fin.  or  finish,  tool,  definition  of,  53,  pl7 
First  degree  equations,  52,  p24 
Fits,  shrinking,  54,  p74 
Fixed,  definition  of,  54,  p6 
Flexibility,  definition  of,  §4,  pG 
Float,  why  some  bodies,  54,  p32 
Fluid,  definition  of,  54,  p2 

ounce,  §1,  pl09 
Foaming,  cause  of,  §4,  p88 
Foci  of  ellipse,  52,  p91 
Focal  length  of  lens,  §4,  pl07 
Focus,  principal,  of  lens,  54,  pl07 
Focussed,  meaning  of,  §4,  pl07 
Foot,  cubic,  definition  of,  §1,  pill 

of  perpendicular,  52,  poo 

square,  definition  of,  51,  pill 
Torce,  definition  of,  54,  p9 

measure  of.  54,  plO 

various  names  of,  54,  p9 
Formula,  definition  of,  52,  pi 

prismoidal,  52,  plOS 
Formulas,  application  of,  52,  p22 
Fraction  and  integer,  product  of,  51.  p58 

common,  to  reduce  a,  to  a  decimal,  {1, 
p74 

complex,  51.  p62 

to  simplify  a,  51.  pG3 

compound,  51,  pG4 

denominator  of,  §1,  p4S 

mixed,  §1,  p75 

numerator  of,  51.  p4S 

power  of,  51.  pG5 

proper  or  improper,  51.  p49 


INDEX 


125 


Fraction,  reduction  of  integer  to  improper) 

§1.  p52 
reduction  of  mixed  Dumber  to  improper, 

SI.  p52 

reduction  of,   to  a  given  denominator, 
51.  p61 

signs  of,  52,  pl6 

terms  of,  §1,  p49 

to  divide  by  a  fraction,  §1,  pGl 

to  divide  by  an  integer,  §1,  pGO 

value  of,  SI.  p49 
Fractions,  addition  of,  rule  for,  §1,  p55 

common  or  vulgar,  §1,  p48 

decimal,  §1,  p<>7 

how  to  write,  $1.  p50 

multiplication  of,  §1,  p58 

reduction  of,  §1,  p50 

reduction  of,  to  common  denominator, 
SI,  p52 

reduction  of,  to  lowest  terms,  §1,  p51 

subtraction  of,  rule  for,  51,  p57 

sum  of  two,  §1,  p55 
Front  elevation,  53,  p6 

view,  53,  p6 
Froth,  cause  of  and  remedies  for,  §■!,  p89 
Frustum  of  cone,  52,  pl22 

of  pyramid,  volume  of,  §2,  plOO 
Full  lines,  use  of,  S3,  pll 
Fusion,  latent  heat  of,  §4,  pSl 


g,  value  of,  §4,  pl3 

Gain  or  loss  per  cent.,  rule  for,  §1,  plOl 

Gallon,   United   States,  wine,  and  imperial, 

SI,  pl09 
Gas,  definition  of,  §1,  p2 

permanent,  definition  of,  §4,  pll 
Gases,  mixture  of,  54,  pip  J 

perfect  and  imperfect,  §4,  p44 

specific  gravity  of,  54,  p35 
Gauge  pressure,  54,  p54 
Gauges,  vacuum,  §4,  p47 
Gay-Lussac's  law,  S4,  p60 
G.  C.  D.,  definition  of,  §1,  p43 
Generatrix,  definition  of,  52,  pi  17 
Geometrical  center  of  polygon,  52,  p66 
Glass,  magnifying,  54,  pl08 
Gram,  definition  of,  §1,  pi  15 
Gravity,  center  of,  52,  pl40 

specific,  definition  of,  54,  i>'i-'i 

specific,  formulas  for,  S4,  pp34,  35 
Great  circle,  52,  pl26 
Greatest  common  divisor,  §1,  p43 
Great  gross,  definition  of,  SI.  pi  13 
Grind,  definition  of,  53,  pi 7 
Gross,  definition  of,  §1,  pi  13 


Hardness,  definition  of,  54,  p6 
Heat  capacity,  54,  p68 


Heat  capacity,  conduction  of,  §4,  p75 
definition  of,  54,  p66 
dynamical  theory  of,  54,  p78 
expansion  of  bodies  by,  54,  p72 
latent  and  sensible,  54,  p81 
mechanical  equivalent  of,  54,  p67 
quantity  of,  S4,  p67 
radiant,  54,  p78 
specific,  54,  p68 
transmission  or  propagation,  54,  p75 

Hectare,  definition  of,  §1,  pll4 

Height  of  arc,  to  find,  52,  p75 

of  half  the  arc,  to  find,  S2,  p76 

Homologous,  definition  of,  52,  pl32 
lines,  properties  of,  pl33-137 

Horizontal  line,  52,  p46;  54,  plO 

Humidity,  relative,  54,  p87 

Hydrometer,  description  of,  54,  p36 
use  of,  54,  p38 

Hydrostatic  machine,  54,  p23 

Hydrostatics,  definition  of,  54,  pl8 

I 

Identical  equations,  52,  pl8 

Image  or  picture,  53,  pi 

Imaginary  quantity,  definition  of,  52,  p25 

Impenetrability,  definition  of,  54,  p4 

Imperial  gallon,  British,  §1,  pl09 

Imperfect  powers,  §1,  p90 

Improper  fraction,  §1,  p49 

fraction,  reduction  of  integer  to,  §1,  p52 
fraction,  reduction  of  mixed  number  to, 
51.  p52 
Impulse  or  blow,  definition  of,  54,  pl8 
Incidence,  angle  of,  54,  pp97,  101 
Incident  ray,  54,  plOl 
Independent  equations,  §2,  pl8 
Indestructibility,  definition  of,  S4,  p5 
Index,  absolute,  of  refraction,  54,  pl03 
of  refraction,  54,  pl03 
of  root,  SI.  p89 
Inertia,  definition  of,  54,  p5 
Inferior  characters,  52,  p22 
Inscribed  angle,  52,  p72 

polygon,  S2,  p89 
Insulators  (of  heat),  54,  p75 
Integer  and  fraction,  product  of,  51,  p58 
definition  of,  §1,  p8 
to  divide  by  a  fraction,  51,  p61 
reduction  of,  to  improper  fraction,  $1, 
p52 
Integral,  definition  of,  §1,  p9 
Intensity  of  light,  54,  p95 
Intercept,  definition  of,  52,  p71 
Intercepted  arc,  52,  p71 
Interior  angles  of  polygon,  52,  p49 
Intersect,  definition  of,  52,  p45 
Invisible  rays,  54,  pill 
Inverted  plan,  53,  p7 
Inverse  ratio,  $1,  p78 
Involution,  51,  p64 


126 


INDEX 


Kilo    or    kilogram,    definition    of,    §1,   pllo 
Kinetic     energy,     definition     of,     J4,     p!6 


Latent  heat,  definition  of,  J4,  p81 

heat  of  fusion,  J4,  pSl 

heat  of  vaporization,  $4-  p81 
Lateral  sides  of  solid,  52,  plOl 
Law,  Boyle's  or  Mariotte's,  £4.  p57 

Boyle's    and    Gay-Lussac's    combined, 
§4,  p63 

Gay-Lussac's  or  Charles',  J4,  p60 

general,  of  all  machines,  54,  p24 

Pascal's,  ?4,  p21 
Laws  governing  radiation,  |4,  p79 
Least  common  denominator,  §1.  p52 

common  multiple,  Jl,  p45 
L.G.M.,  definition  of,  Jl,  p45 
Length  of  circular  arc,  J2,  pS5 
Lens,  definition  of,  §4,  pl04 

faces  of,  J4,  pl06 

focal  length  of,  §r4,  pl07 

kind  of,  §4,  pl06 

principal  focus,  of,  54,  pl07 
Letters,  use  of,  to  designate  lines,  52,  p44 
Light,  brightness  and  intensity  of,  54,  p95 

diffusion  of,  §4,  plOO 

dispersion     or    decomposition     of,     54, 
pllO 

emission  of,  |4,  p91 

rays,   53.  pi:   54.  p91 

rays  always  straight,  54,  p92 

rays,  beam,  cone,  or  pencil  of,  54,  p92 
Light,  nature  of,  §4.  p90 

reflection    from    curved    surfaces,     54, 
p99 

reflection     from    plane    surfaces,     ?4, 
p97 

refraction  of,  54,  plOl 

velocity  of,  54,  p93 

visible  and  invisible  rays  of,  54,  pill 

wave  length  of,  J4.  p91 

waves,   lengths   and   vibrations   of,    54, 
pll5 

white,  54,  pllO 
Like  numbers,  jl,  p2 

terms.   §2,  p5 
Liquid,  definition  of,  J4,  p2 

measure,  table  of,  §1,  pl09 
Linear  equations,  52,  p24 

measure,  table  of,  51,  pl07 
Line,  broken,  52,  p44 

curved,  52,  p44 

horizontal,    vertical,    and    plumb,     §2, 
p46 

mathematical,  52,  p43 

right,  52,  p44 

straight,  definition  of,  52.  p43 


Line,  vortical,  horizontal  or  plumb,  J4,  plO 

of  action,  definition  of,  54,  pl9 
Lines,  broken,  use  of,  §3,  pl2 

broken  and  dotted,  use  of,  53,  pl2 

center,  53,  pplO,  13 

designation  of,  by  letters,  §2,  p44 

dimension,  §3,  pl3 

dotted,  use  of,  §3,  pl2 

full,  use  of,  §3,  pll 

parallel,  52,  p45 

perpendicular,  §2,  p45 

section,  53,  pl9 

thickness  of,  §3,  pll 

weight  of,  §3,  pll 
Liter,  definition  of,  Jl,  pll5 
Long  division,  Jl,  p31 
Longitudinal  section,  J2,  pl02 
Luminous  bodies,  54,  p91 


M 


Machines,  general  law  of  all,  J4,  p24 
Machine,  hydrostatic,  J4,  p23 
Magnifying  glass,  J4,  pl08 
Major  and  minor  axes  of  ellipse,  J2,  p91 
Malleability,  definition  of,  J4,  p6 
Mariotte's  law,  54,  p57 
Mass.  definition  of,  J4,  pll 
Mathematics,  definition  of,  Jl,  p2 
Mathematical  formula,  definition  of,   J2,  pi 

line,  definition  of,  J2,  p43 
Matter,  definition  of,  J4,  pi 

general  properties  of,  54,  pi 

specific  properties  of,  J4,  p6 

three  forms  of,  J4,  pi 
Mean,  arithmetical,  Jl,  pi 25 
Measure,  angular,  table  of,  Jl,  pllO 

circular,  J2,  p83 

cubic,  table  of,  Jl,  pl08 

dry,  table  of,  Jl,  pi 09 

liquid,  table  of,  Jl,  pl09 

linear,  table  of,  Jl,  pl07 

square,  table  of,  Jl,  pl07 

time,  table  of,  Jl,  pll3 
Measurements,  accuracy  in,  52,  p28 

units  of,   Jl,  pi  10 
Measure  of  force,  J4,  plO 
Mechanical  equivalent  of  heat,  J4,  p67 
Medium  (for  light),  definition  of,  54,  p92 
Melting  point,  table  of,  54,  pS6 
Members  of  an  equation,  J2,  pl8 
Meniscus  (lens),  J4,  pl06 

definition  of,  J4,  p41 
Mensuration,  definition  of,  52,  p43 
Mercurial    barometer,    description    of,     J4, 

p52 
Mercury,  weight  of  1  cu.  in.,  54,  p46 
Meter,  definition  of,  Jl,  pll3 
Metric  system,  }1,  ppll3-116 

ton,  Jl,  pll5 
Microscope,  simple  and  compound,  J4,  pl08 


INDEX 


127 


Minuend,  definition  of,  jl,  pi 8 
Minus,  definition  of,  SI,  pl8 
Minutes  (of  arc),  definition  of,  §1,  pi  10 
Mixed  fractions,  §1,  p75 

number,  §1,  p49 

numbers,  how  to  write,  §1,  p50 

numbers,  addition  of,  §1,  p5G 

number,  definition  of,  §1,  p9 

numbers,  division  of,  §1,  p02 

numbers,  multiplication  of,  §1,  p59 

number,     reduction     of,     to     improper 
fraction,  §1,  p52 

number,  to  divide  by  a  mixed  number, 
§1,  p62 

number,   to  divide  by  an   integer,    §1, 
p61 
Mixtures,  temperature  of,  §4,  p70 
Mixture  of  gases,  §4,  pC4 
Mobility,  definition  of,  §4,  po 
Molecule,  definition  of,  §4,  p2 
Money,  United  States,  table  of,  §1,  pi  12 
Monomial,  definition  of,  §2,  p9 
Motion,  definition  of,  §4,  p7 

direction  of,  §4,  p7 
Multiple,  least  common,  §1,  p45 

prime,  §1,  p45 

of  a  number,  §1,  p34 
Multiplicand,  definition  of,  §1,  p22 
Multiplication,  abbreviation  in,  §2,  p30 

check  for,  §1,  p26 

definition  of,  §1,  p22 

rule  for,  §1,  p26 

sign  of,  §1,  p22 

by  a  power  of  10,  §1,  p66 

of  compound  numbers,  §1,  pl22 

of  decimals,  §1,  p69 

of  fractions,  rule  for,  §1,  p59 

of  mixed  numbers,  §1,  p59 

of  monomials,  §2,  p7 

of  polynomials,  §2,  ppll-13 

of  numbers,  §1,  pp24-26 

table,  §1,  p23 
Multiplier,  definition  of,  §1,  p22 


X 


Names  of  periods,  §1,  p7 
Natural  color,  §4,  pill 
Naught,  cipher,  or  zero,  §1,  p5 
Negative  and  positive  quantities,  §2,  p2 
Normal)  definition  of,  §4,  p21 

pressure  on  submerged  surface,  §4,  p22 
Non-luminous  bodies,  §4,  p91 
Notation,  Arabic,  §1,  pp3,  5 

definition  of,  §1,  p3 

Roman,  §1,  pp3,  4 
Notes  on  drawings,  §3,  plC 
Number,  abstract,  §1,  p2 

concrete,  §1,  p2 

definition  of,  §1,  pi 

dividing  into  periods,  {1,  p6 


Number,  mixed,  jl,  p49 

multiple  of,  §1,  p34 

reading  a,  §1,  p7 

root  of  a,  §1,  p89 

significant  part  of,  §2,  p27 

value  represented  by  figures  of  a,  }1,  p6 

composite,  definition  of,  §1,  p35 

denominate,  definition  of,  §1,  plOO 

even,  definition  of,  §1,  p35 

mixed,  definition  of,  §1,  p9 

odd,  definition  of,  §1,  p35 

prime,  definition  of  a,  §1,  p35 

simple    and    compound,    definition   of, 
§1,  pl06 

whole,  definition  of,  §1,  p8 

mixed,  I  eduction  of,  to  improper  frac- 
tion, §1,  p52 

mixed,  to  divide  by  a  mixed  number, 
§1,  p62 

mixed,  to  divide  by  an  integer,  §1,  p61 

of  significant  figures  required,  §2,  p32 
Numbers, 

addition  of,  §1,  ppl3-18 

average  of,  §1,  pi 25 

cube  and  fifth  root  of,  §2,  p33 

divisibility  of,  §1,  p36 

like,  §1,  p2 

division  of,  §1,  p27 

multiplication  of,  §1,  pp24-26 

names  of,  §1,  pp3,  4 

powers  of,  §1,  p64 

properties  of,  §1,  pp34-38 

subtraction  of,  §1,  pl8 

unlike,  §1,  p2 

compound,  addition  of,  §1,  pl20 

compound,  division  of,  §1,  pl23 

compound,  multiplication  of,  §1,  pl22 

compound,  reduction  of,  §1,  ppl  10-120 

compound,  subtraction  of,  §1,  pl21 

mixed,  addition  of,  §1,  p56 

mixed,  division  of,  §1,  pG2 

mixed,  multiplication  of,  §1,  p59 

mixed,  subtraction  of,  §1,  p57 
Numeration,  definition  of,  §1,  p3 
Numerator  of  a  fraction,  §1,  p48 
Numerical  operations,  accuracy  in,  §2,  p29 


Obtuse  angle,  §2,  p4G 

Odd  number,  definition  of,  §1,  p35 

Opaque  body,  §4,  p92 

Open  belt,  to  find  length  of,  §2,  p80 

Optic  center,  §4,  100 

Order  of  a  figure,  §1,  pO 

of  a  figure,  higher  or  lower,  §1,  p9 
Ordinate,  definition  of,  §2,  p94 
Orthographic  projection,  §3,  p2 
Ounce,  fluid,  §1,  pl09 
Over-all  dimensions,  §3,  pl4 


128 


INDEX 


Paper,  coloring,  reason  for,   §4,  pi  IS 
Parallel  lines,  §2,  p45 

Parallelogram,  altitude  and  diagonal  of,  |2, 
p61 

area  and  base  of,  §2,  p62 

definition  of,  §2,  p61 
Parallelograms,  properties  of.  §2,  pf>l 
Parallelopiped,  definition  of,  §2,  102 
Partial  vacuum,  definition  of,  §4,  p47 
Pascal's  law,  §4,  p21 
Path  of  a  body,  §4,  p7 
Pencil  of  light  rays,  §4,  p92 
Penumbra,  definition  of,  §4,  p95 
Per  cent,  definition  of,  §1,  p9S 

gain  or  loss,  rule  for,  §1,  pl04 
Percentage,  definition  of,  §1,  p98 

rule  to  find,  §1,  100 
Perch,  definition  of,  §1,  plOS 
Perfect  gas,  definition  of,  §4,  p44 

powers,   §1,  p90 

vacuum,  definition  of,  54,  p47 
Perimeter,  definition  of,  §2,  poS 

of  polygon,  §2,  p48 

of  triangle,  §2,  p58 
Periods,  dividing  a  number  into,  §1 ,  p6 

names  of,  §1,  p7 
Periphery,  definition  of,  §2,  p70 
Permanent  gas,  definition  of,   §4.  p44 
Perpendicular,  foot  of,  §2,  poo 

lines,  §2,  p45 
Picture  or  image,  §3,  pi 

plane,  the,  §3,  pi 
Pigment,  definition  of,  §4,  pllo 
Pigments,  mixing,  §4,  pi  15 
Pipe,  volume  of,  §2,  pll5 
Piston,  stroke  of,  §4,  p49 
Photometers,    description    of,    §4,    pp95-97 
Plan,  §3,  p6 

inverted,  §3,  p7 
Plane,  definition  of,  §2,  p46 

the  picture,  §3,  pi 

trace  of,  §3,  p4 

of  symmetry,  §2,  pl39 

figure,  area  of  any,  §2,  p93 

figure,  definition  of,  §2,  p48 

figures,  projection  of,  §3,  p25 
Plano-convex  lens,  §4,  pl06 
Plano-concave  lens,  §4,  plOC 
Planed,  definition  of,  §3,  pl7 
Pliability,  definition  of,  §4,  p6 
Plumb  line,  §2,  p46;  §4,  plO 
Plus,  definition  of,  §1,  pll 

or  minus  sign,  §2,  p24 
Pneumatics,  definition  of,  §4,  p42 
Point,  decimal,  definition  of,  §1,  p8 

dew,  §4,  pS7 

of  application,  54,  pl4 

of  intersection,  §2,  p44 

of  sight,  §3,  p2 


Point  of  tangency,  §2,  p71 

projection  of,  §3,  p25 
Pole,  definition  of,  §2,  pi 24 
Polyedron,  definition  of,  §2,  pl02 
Polygon,  convex,   §2,  p66 

definition  of,  §2,  p48  v 

geometrical  center  of,  §2,  p66 

inscribed  and  circumscribed,  52,  pS9 

interior  angles  of,  §2,  p49 

perimeter  of,  §2,  p4S 

re-entrant  or  concave;  §2,  pGO 

regular,  §2,  p49 

sides  and  angles  of,  §2,  p4S 
Polynomial,    arranging,    according    to    de- 
scending powers,  §2,  p9 

definition  of,  §2,  p9 
Polynomials,  addition  of,  §2,  plO 

division  of,  §2,  ppl3-15 

multiplication  of,  §2,  ppll-13 

subtraction  of,  §2,  plO 
Polygons,  area  of  regular,  §2,  p68 

names  of,  §2,  p49 

properties  of  regulai,  52,  pCC 

table  of  regular,  §2,  p68 
Porosity,  definition  of,  54,  p3 
Positive  and  negative  quantities,  52,  p2 
Potential  energy,  definition  of,  §4,  pl6 
Power  of  a  fraction,  §1,  p65 
Powers  of  10,  §1,  p65 

of  numbers,  §1 ,  pC4 

perfect  and  imperfect,  §1 ,  p90 
Pressure,  absolute,  54,  p53 

absolute  zero  of,  §4,  p54 

at  any  depth,  §4,  p26 

definition  of,  54,  plS 

deptli  required  for  a  given,  §4,  p2G 

due  to  weight  of  liquid,  5-4.  p25 

from  vacuum  gauge  reading,  §4,  p47 

gauge,  54,  p54 

normal,  on  submerged  surface,   54,  p22 

of  atmosphere,  54,  p46 

on  submerged  surface  due  to  weight  of 
liquid,  §4,  p27 

specific  and  total,  54,  pl9 
Pressures,  combined,  on  submerged  surface, 

§4,  p29 
Primary  colors,  54,  pill 
Prime  factors,  §1,  p36 

marks,  52,  p21 

multiple,  51.  p45 

or  prime  number,   definition   of  a,    51. 
p35 
Priming,  cause  of  and  prevention  of,  54,  p90 
Principal  axis  of  lens,  §4,  plOG 

focus  of  lens,  §4,  pl07 
Prism,  altitude  of,  52,  pi 02 

area  and  volume  of,  52,  pl03 

definition  of,  52,  pl02 

refraction  through,  §4,  pi 04 
Prismatic  colors,  54,  pl09 
Prismatoids  and  prismoids,  §2,  pl07 


1ND10X 


129 


l'i ismoidal  formula,  $2,  plOS 
Projection  drawing,  §•'!,  p2 

of  a  cylinder,  S3,  p25 

of  a  point,  §3,  p25 

of  plane  figures,  §3,  p25 

orthographic,  §3,  p2 
Product,  definition  of,  §1,  p22 
Projection,  definition  of,  §2,  p58 

Boenographio,  §3,  p2 
Projectors,  §3,  p4 
Propagation  of  heat,  §1,  p75 
Proper  Fraction,  §1,  p49 
Proportion,  compound,  §1.  p83 

dclinilion  of,   §1,  p79 

direct  and  inverse,  §1,  p80 

how  to  read  a,  §1,  p80 

law  of,  jl,  p80 

names  of  terms  of  a,  §1,  p80 

three  ways  of  writing,   §1,  p80 

to  find  value  of  unknown  term,  §1,  p81 
Protractor,  $2,  p46 
Pull,  definition  of,  §4,  p9 
Pulley,  crown  of,  §3,  p23 
Pump,  suction,  description  of,  j4,  p48 

work  required  to  operate,  §4,  p49 
Push,  definition  of,  §4,  p9 
Pyramid,  area  and  volume  of,  §2,  pl05 

axis  of,  §2,  pl05 

definition  of,  §2,  pl04 

frustum  of,  volume  of,  §2,  pl06 

regular,   52,  pl05 

slant  height  of,  §2,  pl0.5 

vertex  and  altitude  of,  §2,  pl04 

Q 

Quadrant,     defintion     of,      §1,     ppllO,  112 
Quadratic   equation,   formula    for   roots   of, 
§2,  p25 

equations,  §2,  p23,  24 

equations,     solution     of,     §2,     p25 
Quantity,  definition  of,   §1,  p2 

imaginary,  definition  of,  §2,  p25 

literal,  §2,  p3 

numerical,  §2,  p3 

numerical  value  of,   §2,  p3 

of  heat,  §4,  p67 

unknown,  §2,  p23 
Quantities,  positive  and  negative,  §2,  p2 
Quire,  definition  of,  jl,  pi  13 
Quotient,  definition  of,  §1,  p28 

R 

R,  value  of  for  gas  or  air,  §4,  pp63,  04 
Radial  section,  §3,  p23 

section,  definition  of,  $2,  pl43 
Radian,  definition  of,  §2,  p84 
Radiant,  definition  of,  §4,  p93 

heat,  §4,  p78 
Radiation,  definition  of,  §4,  p77 

laws  governing,  §4,  p79 
9 


Radius,  definition  of,  §2,  p70 

of  arc,  to  find,  §2,  p75 

of  circle,  to  find,  §2,  p77 

of  sphere,  §2,  pi 24 
Rate  and  rate  per  cent.,  difference  between, 
§1,  p99 

per     cent.,     definition     of,      §1,     p99 

rule  for  equivalent  discount,  §1,  pl03 

rule  to  find,  §1,  plOO 
Ratio,  definition  of,  §1,  p78 

direct  and  inverse,  §1,  p78 

signs  of,  §1,  p78 

terms  of,  §1,  p78 

value  of,  §1,  p78 
Ray,  incident,  §4,  plOl 

of  light,  §4,  p91 

refracted,  §4,  plOl 
Rays,  actinic,  §4,  p91 

definition  of,  §2,  pl31 

light,   §3,  pi 

of   light,  beam,  cone  or  pencil  of,    §4, 

p92 
refrangible,  §4,  pl04 
ultra-red  and  ultra-violet,  §4,  pill 
visible  and  invisible,  §4,  pill 
Reading  a  number,  §1,  p7 
drawings, 

guard  for  movable  jaw  of  vise,  §3,  p28 
mercer  cell,  the,  §3,  p30 
rotary  drying  furnace,  §3,  p33 
worm  washer,  §3,  p33 
drawings,  directions  for,  §3,  pp25-28 
Ream,  definition  of,  §1,  pi  13 

or  reamed,  definition  of,  §3,  pi 6 
Rear  elevation,  §3,  p7 

view,  §3,  p7 
Reciprocal,  definition  of,  §2,  p4 
Reduction    ascending    and    descending,    of 
fractions,  §1.  p51 
of  common  fraction  to  a  decimal,    §1, 

p74 
of     compound  numbers,  §1,  ppl  16-120 
of  decimals  to  common   fractions,    §1, 

pp67,  68 
of  decimals  to  fractions  having  a  given 

denominator,  §1,  p75 
of  fractions,  §1,  p50 
of   fractions  to   common   denominator, 

§1,  p52 
of     fractions     to     lowest     terms,      §1, 

p51 
of  integer  to  improper  fraction,  §1,  p52 
of  mixed  number  to  improper  fraction, 
§1,  p52 
Re-entrant  angle,  §2,  pOO 

polygon,  §2,  p66 
Reflection,  angle  of,  §4,  p97 
of  light,  §4,  pp97-100 
total,  §4.  pi 03 
Refracted  ray,  §4,  plOl 
Refraction,  angle  of,  §4,  plOl 


130 


INDE  X 


Refraction,  index  of,  54,  pl03 

of  light,  54.  plOl 

through  prism,  54,  pl04 
Refrangible  rays,  §4,  pl04 
Regular  polygon,  definition  of,  52,  p49 

polygons,  properties  of,  52,  p66 

polygons,  table  of,  §2,  p68 

pyramid,  §2,  plOo 
Relati%'e,  definition  of,  §4,  p7 
Remainder,  definition  of,  §1,  plS,  28 
Rest,  definition  of,  54,  p6 
Rhomboid,  definition  of,  §2,  p61 
Rhombus,  definition  of,  §2,  p61 
Right  angle,  definition  of,    §1,  ppllO,   112 
§2,  46 

cone,  §2,  pl20 

cylinder,  §2,  pi  10 

line,  §2,  p44 

parallelopiped,  §2,  pl02 

prism,  §2,  pl02 

section,  §2,  pl02 

triangle,  properties  of,  §2,  p53 
Rigidity,  definition  of,  §4,  p6 
Ring,  area  of,  §2,  p82 
Roman  notation,  §1,  pp3,  4 
Root,  index  of,  §1,  p89 

of  a  number,  $1,  p89 

rule  for  square,  51,  p95 
Roots  of  an  equation,  $2,  p25 
Rules,  trapezoidal  and  Simpson's,  §2,  p95 
Rumford's   method  of  measuring  intensity 
of  light,    §4,  p96 


Satisfied,  when  equation  is  said  to  be,    §2, 

pl9 
Saturated  vapor,  §4,  pS3 
Scale,  Beaum£,  §4,  p37 

drawing,  definition  of,  §3,  plo 

drawing  to,  §3,  pl4 
Scales,  centigrade  and  Fahrenheit,  $4,  p43 

definition  of,  §3,  pl4 

special,  §3,  pl6 
Scaling  a  drawing,  §3,  pi 6 
Scenographic  projection,  §3,  p2 
Scraped,  definition  of,  §3,  pl7 
Secant,  definition  of,  |2,  p70 
Second  degree  equations,  §2,  p24 
Secondary  colors,  §4,  pill 
Seconds  (of  arc),  definition  of,  51,  pllO 
Section,  cross,  §3,  p23 

lines,  §3,  pl9 

radial,  53,  p23 

radial,  definition  of,  52,  pl43 
Sectional  view,  53,  pl7 
Sections,  conventional,  53,  p22 

definition  of,  53,  pl7 

right,  cross,  and  longitudinal,  52,  pl02 

standard,  53,  ppl9-21 

thin,  53,  p21 


Sector,  area  of,  |2,  p87 

definition  of,  52,  p70 

spherical,  base  of,  52,  pl27 
Segment,  area  of,  52,  p87 

of  circle,  definition  of,  52,  p70 

of  line,  definition  of,  52,  p43 

spherical,  52,  pl26 

spherical,  volume  of,  52,  pl26 
Semicircle,  definition  of,  52,  p70 
Semi-transparent  body,  54,  p92 
Sensible  heat,  definition  of,  54,  p81 
Shadow,  definition  of,  54,  p94 
Short  division,  §1,  p29 
Shrinking  fits,  5-4,  p74 
Side  elevation,  §3,  p6 

of  square,  length  of,  52,  p64 

view,  53,  p6 
Sides  of  a  solid,  52,  plOl 

of  angle,  52,  p45 
Sight,  point  of,  53,  p2 
Sign,  double,  meaning  of,  52,  p24 

of  equality,  definition  of,  $1,  pll 

of  multiplication,  §1,  p22 

of  product  of  two  monomials,  52,  p7 

plus,  definition  of,  §1,  pll 

plus  or  minus,  52,  p24 

of  quotient,  52,  p8 

of  subtraction,  §1,  pl8 
Significant  figures,  52,  p27 

figures,  number  of,  required,  52,  p32 

part  of  a  number,  §2,  p27 
Signs  of  a  fraction,  52,  pl6 

of  aggregation,  §1,  p76;  52,  pl5 

of  division,  §1,  p28 

of  ratio,  §1,  p78 
Similar  figures,  52,  pl31 

triangles,  52,  p52 
Simple  microscope,  54,  pl08 

number,  definition  of,  §1,  pl06 
Simpson's  rule,  52,  p95 
Sink,  why  some  bodies,  54,  p32 
Siphon,  description  of,  54,  p50 
Slant  height  of  cone,  52,  pl21 

height  of  pyramid,  52,  plOo 
Small  circle,   52,  pl25 
Solar  spectrum,  54,  pl09 
Solid,  definition  of,  §4,  p2 

of  revolution,  volume  of,  52,  pl42 

sides,  bases,  and  edges  of,  52,  plOl 
Specific    gravity    and    density,    distinction 
between,  54,  p35 

gravity  by  hydrometer,  54,  p37 

gravity,  definition  of,  54,  p33 
formulas  for,  54,  pp34,  35 

gravity  of  gases,  54,  p35 

heat,  definition  of,  §4,  p68 
table  of,  54,  pp69,  70 

pressure,  definition  of,  54,  pl9 

weight,  54,  pl4 
Spectroscope,  description  of,  54,  pi  13 
Spectrum  analysis,  54,  pi  12 


INDKX 


131 


Spectrum,  solar,  §4,  pl()9 

Sphere,  area  and  volume  of,  §2,  pl25 

axis  of,  §2,  pi 24 

center  of,   §2,  pl24 

definition  of,  §2,  pl24 

radius  and  diameter  of,  §2,  pl24 
Spherical  sector,  base  of,  §2,  pi 27 

segment,  definition  of,  §2,  pl26 

segment,  volume  of,  §2,  pl26 

surface,  §2,  pl24 
Square,  diagonal  of,  length  of,  §2,  p63 

foot,  definition  of,  §1,  pill 

measure,  table  of,  §1,  p!07 

root,  rule  for,  §1,  p95 

side  of,  length  of,  §2,  p64 
Standard  candle,  §4,  p95 

conditions,  definition  of,  §4,  p36 

sections,  J3,  ppl9-21 
Steam,  dry,  saturated,  or  superheated,   §4, 
p83 

temperature  of  dry  and  saturated,   §4, 
p83 

total  heat  of,  §4,  p84 
Straight  line,  definition  of,  §2,  p43 
Stroke  of  piston  or  plunger,  §4,  p49 
Subscripts,  §2,  p22 
Subtended,  definition  of,  §2,  p71 
Subtraction,  definition  of,  Jl,  pl8 

of  compound  numbers,  §1,  pl21 

of  decimals,  §1,  p68 

of  fractions,  §1,  p56 

of  monomials,  §2,  p6 

of  polynomials,  §2,  plO 

rule  for,  §1,  pl9 

sign  of,  §1,  pl8 
Subtrahend,  definition  of,  §1,  pl8 
Suction,  height  of,  §4,  p49 

pump,  description  of,  §4,  p48 
Sum,  definition  of,  §1,  pll 

of  two  fractions,  §1,  p55 
Superheated  vapor,  §4,  p83 
Superior  characters,  §2,  p22 
Superposition,  §2,  p52 
Surface,  conical,  §2,  pi 20 

cylindrical)  §2,  pi  10 

definition  of,  §2,  p47 

spherical,  §2,  pl24 

tension,  §4,  p88 
Symmetrical  figures,  §2,  pl37 
Symmetry,  axis  of,  §2,  pl37 

center  of,  §2,  pl38 

plane  of  §2,  pl39 

with  respect  to  a  point,  §2,  pl40 


Table,  addition,  §1,  pl2 

multiplication,  §1,  p23 

of  coefficients  of  conductivity,  §4,  p76 

of  coefficients  of  expansion,  §4,  p73 


Table    of    melting    and   boiling   points,   §4, 
p86 

of  specific  heats,  §4,  pp69,  70 
Tangency,  point  of,  §2,  p71 
Tangent,  definition  of,  §2,  p71 
Tap,  definition  of,  §3,  pl7 
Teaspoon,  definition  of,  §1,  pi  15 
Temperature,  absolute,  §4,  pCl 

absolute  zero  of,  §4,  p61 

of  mixtures,  §4,  p70 
Tenacity,  definition  of,  §4,  p6 
Tension  of  gases,  definition  of,  §4,  p42 

surface,  §4,  p88 
Terms,  like  and  unlike,  §2,  p5 

of  a  fraction,  §1,  p49 

of  a  ratio,  §1,  p78 
Tertiary  colors,  §4,  pill 
thds.,  definition  of,  §3,  pl6 
Thermal  conductivity,  definition  of,  §4,  p76 

unit,  British,  §4,  p67 
Thermometer,  definition  of,  §4,  p42 
Thickness  of  lines,  §3,  pll 
Three-color  process,  §4,  pi  10 
Time,  measures  of,  table  of,  Jl,  pi  13 
Ton,  long  and  short,  difference  between,  Jl, 

pl08 
Tonne  or  metric  ton,  Jl,  pi  15 
Tool  fin.  or  tool  finish,  definition  of,  J3,  pl7 
Top  view,  J3,  p6 
Torrecellian  vacuum,  §4,  p46 
Torus,  volume  of,  §2,  pl42 
Total  energy,  J4,  pl7 

pressure,  definition  of,  §4,  pl9 

reflection,  J4,  pl03 
Trace,  front,  J3,  p4 

of  a  plane,  J3,  p4 

side,  §3,  p4 
Transformation  of  equations,  §2,  ppl9,  21 
Translucent  body,  §4,  p92 
Transmission  of  heat,  J4,  p75 
Transparent  body,  J4,  p92 
Transposing  terms  of  equation,  §2,  pl9 
Trapezium,  area  of,  J2,  p65 

definition  of,  J2,  p64 
Trapezoid,  area  of,  J2,  p64 

definition  of,  J2,  p64 
Trapezoidal  rule,  J2,  p95 
Trial  dividend,  Jl,  p92 

divisor,  Jl,  p92 
Triangle,  altitude  of,  J2,  p56 

base  of,  J2,  p51 

perimeter  of,  J2,  p58 

relation  between  sides  of,  J2,  p59 
Triangles,  area  of,  J2,  pp55-58 

classification  of,  J2,  p50 

properties  of,  J2,  p51 

proportionality  of,  J2,  p52 

right,  properties  of,  J2,  p53 

similar,  J2,  p52 
Troy  weight,  Jl,  pl09 
Tube,  volume  of,  J2,  pi  15 


132 


INDEX 


u 


Ultra-red  and  ultra-violet  rays,  §4,  pill 
Umbra,  definition  of,  §4,  p95 
I'liguhi.  area  of,  §2,  pi  13 

volume  of,  §2,  pi  14 
Uniform  velocity,  §4,  p8 
Unit,  British  thermal,  §4,  p67 

definition  of,  §1,  pi 

of  acceleration,  §4,  p9 

of  work,  §4,  pl4 
United  States  gallon,  §1,  pl09 

money,  table  of,  §1,  pi  12 
Units  of  measurement,  §1,  pi  10 
Unknown  quantity,  definition  of,  §2,  p23 
Unlike  numbers,  §1,  p2 

terms,  §2,  p5 


Vacuum,  definition  of,  §1,  p33 

gauges,  §4,  pp47,  54 

perfect  and  partial,  §4,  p47 

Torricellian,  §4,  p46 

weight  in,   §4,  p33 
Value  of  a  fraction,  §1,  p49 

of  ratio,  §1,  p78 
Vaporization,  latent  heat  of,  §4,  pSl 
Variable  velocity,  §4,  pS 
Variables,  definition  of,  §2,  p32 
Velocity,  definition  of,  §4,  p7 

of  light,   §4,  p93 

uniform  or  constant,  §4,  pS 

unit  of,   §4,  p8 

variable,   §4,  pS 
Vertex  of  angle,  §2.  p45 

of  cone,  §2,  pl20 

of  pyramid,  §2,  pl04 
Ycrtcxes  of  ellipse,  §2,  p91 
Vertical  line,  §2,  p46 

line,  definition  of,  §4,  plO 
Vibgyor,  meaning  of,  §4,  pi 09 
View,  back  or  rear,  §3,  p7 

bottom,  §3,  p7 

front,  §3,  p6 

sectional,  §3,  pl7 

side,  §3,  p6    , 

top,  §3,  pG 
Views,  names  of,  §3,  p6 

number  of,  §3,  pG 
Viscosity,  definition  of,  §4,  p88 


Visible  rays,  54.  pill 

Visibility  of  objects,  §4,  plOl 

Visualizing  objects  from  drawings,  §3,  p23 

Volume  of  any  cylinder,  §2,  pi  14 
of  cone,  §2,  pl21 

of  cylindrical  ring  or  torus,  §2,  pl42 
of  frustum  of  cone,  §2,  pl23 
of  pipe  or  tube,  §2,  pi  15 
of  prism,   §2,  pl03 
of  pyramid,  §2,  pl05 
of  sphere,  §2,  pl25 
of  spherical  sector,  §2,  pi 27 
of  spherical  segment,  §2,  pl26 
of  solid  of  revolution,  §2,  pl42 
of  ungula,  §2,  pll4 
of  wedge,  §2,  pl09 


W 


Water  seeks  its  level,  explanation  of,  §4,  p2S 

Wave  length  of  light,  §4,  p91 

Waves,  light,  lengths  and  vibrations  of,  §4, 

pll5 
Wedge,  volume  of,  §2,  pl09 
Weight,  avoirdupois,  table  of,  §1,  pl08 

definition  of,  §4.  p5 

in  vacuum,  §4,  p33 

of  air  or  gas,  formula  for,  §4,  p64 

of  lines,  §3,  pll 

specific,  §4,  pl4 

troy,  §1,  pl09 

variation  of,  §4,  pl3 
White  light,  §4,  pi  10 
Whole  number,  definition  of,  §1,  p8 
Winchester  bushel,  §1,  pl09 
Wine  gallon,  §1,  pl09 
Work,  definition  of,  §4,  pl5 

required  to  operate  pump,  §4,  p49 

unit  of,  §4,  pl4 
Working  drawing,  §3,  p8 


Yard,  definition  of,   §1,  pi  10 
Year,  length  of,  §1,  pi  13 


Zero,  absolute,  of  pressure,  §4,  p54 
absolute,  of  temperature.  §4,  p61 
naught,  or  cipher,  $1,  p5 

Zone,  area  of,  §2,  pl26 


~*   ,+ure  of  ?ulP  *  Paper 
T^e  Hanuf  actur  ng 

1   .  „«  -Physics 


